$$
\leqalignno{
&\partial\Vert v\Vert_{h_{V,\Delta}}^2=
\langle\!\langle\partial v,v\rangle\!\rangle_{h_{V,\Delta}}+\sum_j
\cr
&\qquad{}+|\sigma_j|^{-2+2/\rho_j}\Big(\partial^2\sigma_j(dz,v)+(-1+1/\rho_j)\,
\sigma_j^{-1}\partial\sigma_j(dz)\,\partial\sigma_j(v)\Big)\,
\overline{\partial\sigma_j(v)}
&(5.13_3)\cr
&\qquad{}-{1\over\rho_j}\,|\sigma_j|^{-2+2/\rho_j}\,
\partial\sigma_j(v)\,\ddbar\varphi_j(dz,v)
&(5.13_5)\cr
&\qquad{}+K'\,|\sigma_j|^{-2+2/\rho_j}|v|^2\,\overline\sigma_j\,
\partial\sigma_j(dz)
&(5.13_6)\cr}
$$


Denoting by $\alpha^*$ the adjoint vector to a $1$-form $\alpha$,
so that $\alpha=\langle\bu,\alpha^*\rangle$, this can be rewritten
$$
\leqalignno{
\big\langle\sum_j
\big(|\sigma_j|^{-2+2/\rho_j}
&\partial\sigma_j^*\otimes\partial\sigma_j
+K+K'\,|\sigma_j|^{2/\rho_j}\Id\big)\,\Gamma(\xi)\cdot v,w\big\rangle\cr
={}&\bigg\langle\sum_j|\sigma_j|^{-2+2/\rho_j}\,
\Big(\partial^2\sigma_j(\xi,\bu)+(-1+1/\rho_j)\,\sigma_j^{-1}
\partial\sigma_j(\xi)\,\partial\sigma_j\Big)(v)\otimes\partial\sigma_j^*
&(Z_0)\cr
&-|\sigma_j|^{-2+2/\rho_j}\,\partial\sigma_j(v)\,
\overline\sigma_j\otimes\ddbar\varphi_j(\bu,\xi)^*
&(Z_1)\cr
&+K'\,{1\over\rho_j}\,|\sigma_j|^{-2+2/\rho_j}\,
\partial\sigma_j(\xi)\,\overline\sigma_j\,v,w\bigg\rangle.
&(Z_2)\cr}
$$
Therefore we find
$$
\eqalign{
\Gamma(\xi)\cdot v={}&
\bigg(\sum_j
\big(|\sigma_j|^{-2+2/\rho_j}\,\partial\sigma_j\otimes\partial\sigma_j^*
+K+K'\,|\sigma_j|^{2/\rho_j}\Id\big)\bigg)^{-1}\cdot\cr  
&\quad{}\bigg(\sum_j|\sigma_j|^{-2+2/\rho_j}
\Big(\partial^2\sigma_j(\xi,\bu)+(-1+1/\rho_j)\,
\sigma_j^{-1}\partial\sigma_j(\xi)\,\partial\sigma_j(v)\Big)\otimes
\partial\sigma_j^*\cr
&\quad{}-|\sigma_j|^{-2+2/\rho_j}\,
\partial\sigma_j(v)\,\overline\sigma_j\otimes\ddbar\varphi_j(\bu,\xi)^*\cr
&\quad{}+K'\,{1\over\rho_j}\,|\sigma_j|^{-2+2/\rho_j}\,
\partial\sigma_j(\xi)\,\overline\sigma_j\,v\bigg).\cr}
$$

\noindent


As a consequence
$$
\leqalignno{
\Gamma(\xi)\cdot v&+\sum_j\varepsilon\,|\sigma_j|^{-2+2/\rho_j}
\partial\sigma_j(\Gamma(\xi)\cdot v)\,(\partial\sigma_j)^*\cr
={}&\sum_j\varepsilon\,
|\sigma_j|^{-2+2/\rho_j}\Big(\partial^2\sigma_j(\xi,v)+(-1+1/\rho_j)\,
\sigma_j^{-1}\partial\sigma_j(\xi)\,\partial\sigma_j(v)\Big)\,
(\partial\sigma_j)^*
&(5.17_0)\cr
&\quad{}-\varepsilon\,|\sigma_j|^{-2+2/\rho_j}\,
\partial\sigma_j(v)\;\overline\sigma_j\,(\ddbar\varphi_j(\bu,\xi))^*
&(5.17_1)\cr}
$$
where $\alpha^*\in V$ is the dual vector to a $1$-form $\alpha$, such that
$\langle\alpha^*,\bu\rangle=\overline\alpha$. An application of the
linear form $\partial\sigma_\ell$
to $(5.17_0,~(5.17_1)$ implies
$$
\leqalignno{
\partial\sigma_\ell(\Gamma(\xi)\cdot v)\,&
+\sum_j\varepsilon\,|\sigma_j|^{-2+2/\rho_j}
\langle \partial\sigma_\ell,\partial\sigma_j\rangle\,
\partial\sigma_j(\Gamma(\xi)\cdot v)\cr
={}&\sum_j \varepsilon\,
|\sigma_j|^{-2+2/\rho_j}\Big(\partial^2\sigma_j(\xi,v)+(-1+1/\rho_j)\,
\sigma_j^{-1}\partial\sigma_j(\xi)\,\partial\sigma_j(v)\Big)\,
\langle\partial\sigma_\ell,\partial\sigma_j\rangle
&(5.17_2)\cr
&\quad{}-\varepsilon\,|\sigma_j|^{-2+2/\rho_j}\,
\overline\sigma_j\,\partial\sigma_j(v)\,
\langle\partial\sigma_\ell,\ddbar\varphi_j(\bu,\xi)\rangle.
&(5.17_3)\cr}
$$
After multiplying by $|\sigma_\ell|^{-2+2/\rho_\ell}$, the left hand side
can be seen as an expression of the image of the column vector
$(|\sigma_j|^{-2+2/\rho_j}\partial\sigma_j(\Gamma(\xi)\cdot v))_j$
by the hermitian matrix
$(\Id+\varepsilon\langle\partial\sigma_\ell,\partial\sigma_j\rangle)$ which
dominates the identity by a small amount.
We have bounds $|\ddbar\varphi_j(\xi,v)|\le C|\xi||v|$, 
$|\partial\sigma_j(\xi)|\le C|\xi|$, and similarly
$|\partial^2\sigma_j(\xi,v)|\le C|\xi||v|$, for some constant $C>0$.
This implies
$$
\leqalignno{
\sum |\sigma_\ell&|^{-2+2/\rho_\ell}\,\big|\partial\sigma_\ell
(\Gamma(\xi)\cdot v)\big|\le C'\varepsilon\bigg(\sum_\ell|\sigma_\ell|^{-2+2/\rho_\ell}\bigg)\times\cr
&\sum_j|\sigma_j|^{-2+2/\rho_j}\bigg(|\xi|\,|v|+
|\sigma_j|^{-1}\,|\partial\sigma_j(\xi)|\,
|\partial\sigma_j(v)|+|\sigma_j|\,|\xi|\,|\partial\sigma_j(v)|\bigg).
&(5.17_4)\cr}
$$

Since $\overline\sigma_j\,\ddbar\varphi_j(\xi,\eta)=
\langle\xi,\sigma_j\,\ddbar\varphi_j(\bu,\eta)^*\rangle$,
the identity $(5.19_*)$ can be rewritten
$$
\leqalignno{
Q(z,v)&(\xi,\eta)^2\cr
\noalign{\vskip6pt}
\kern40pt&\kern-6pt{}=\gamma\,\ddbar\psi(\xi,\xi)\,|v|^2
+\sum_{\ell,m,\lambda,\mu}c_{\ell m\lambda\mu}\,\xi_\ell\overline\xi_m\,
v_\lambda\overline v_\mu
&(5.20_0)\cr
&+\sum_j\varepsilon\,\bigg(-{1\over \rho_j}\,\ddbar\varphi_j(\xi,\xi)+
\gamma\,\ddbar\psi(\xi,\xi)\bigg)
\,|\sigma_j|^{-2+2/\rho_j}|\partial\sigma_j(v)|^2
&(5.20_1)\cr
&+|\eta|^2+\sum_j\varepsilon\,|\sigma_j|^{-2+2/\rho_j}\,
\Big|\big(\partial\sigma_j(\eta)
+\partial^2\sigma_j(\xi,v)\big)+(-1+1/\rho_j)\,
\sigma_j^{-1}\partial\sigma_j(\xi)\,\partial\sigma_j(v)\Big|^2
&(5.20_2)\cr
&\kern6pt{}+\varepsilon\,|\sigma_j|^{2/\rho_j}\,
\bigg|\delta^{-1/2}\,|\sigma_j|^{-1/\rho_j}\,\ddbar\varphi_j(v,\xi)
-{\delta^{1/2}\over\rho_j}\,|\sigma_j|^{-2+1/\rho_j}\;
\overline{\partial\sigma_j(\xi)}\,
\partial\sigma_j(v)\bigg|^2
&(5.20_3)\cr
&\kern6pt{}-{\delta\,\varepsilon\over\rho_j^2}\,|\sigma_j|^{-4+4/\rho_j}
\,|\partial\sigma_j(\xi)|^2\,|\partial\sigma_j(v)|^2
&(5.20_4)\cr
&\kern6pt{}-\varepsilon\,\big(\delta^{-1}-|\sigma_j|^{2/\rho_j}\big)\,
\big|\ddbar\varphi_j(v,\xi)\big|^2
&(5.20_5)\cr
&\kern6pt{}+\varepsilon\,|\sigma_j|^{-2+2/\rho_j}\,
\bigg|\delta^{-1/2}\,\sigma_j\,\ddbar\varphi_j(\bu,\eta)^*
-\delta^{1/2}\,\partial\sigma_j(v)\,\xi\bigg|^2
&(5.20_6)\cr
&\kern6pt{}-\delta\,\varepsilon\,|\sigma_j|^{-2+2/\rho_j}\,|\xi|^2\,
|\partial\sigma_j(v)|^2
&(5.20_7)\cr
&\kern6pt{}-\delta^{-1}\,\varepsilon\,|\sigma_j|^{2/\rho_j}\,
\big|\ddbar\varphi_j(\eta,\bu)\big|^2
&(5.20_8)\cr}
$$


We prove the lower bound by induction on $r$. For $r=1$,
we have $\ell_j(u)=\alpha_ju_1$ with $\alpha_j=1$ and it is obvious that
one can take $m'_{1,p}=1$. Let us now assume $r\ge 2$.
We write
$$
\eqalign{
I(\ell_1,\ldots,\ell_r)
&=\int_{\bS^{2r-1}}\prod_{j=1}^p|\ell_j(u')+\ell_j(e_r)u_r|^2\,d\mu(u)\cr
&=\int_{\bS^{2r-1}}\prod_{j=1}^p|\ell_j(e_r)|^2~
\Bigg|\sum_{j=0}^pu_r^{p-j}s_j(\ell'_\bu(u'))\Bigg|^2d\mu(u)\cr}
$$
where $\ell'_j(u')=\ell_j(u')/\ell_j(e_r)$ and the $s_j(\bu)$ denote the 
elementary symmetric functions. In the last equality, we make a change
of variable $u_r\mapsto e^{i\theta}u_r$ and take the average on
$\theta\in[0,2\pi]$ by means of Parseval's equality. By (6.9) we get
$$
\eqalign{
I(\ell_1,\ldots,\ell_p)
&=\int_{\bS^{2r-1}}\prod_{j=1}^{r-1}|\ell_j(e_j)|^2\times\sum_{j=0}^p
|u_r|^{2(p-j)}\,\big|s_j(\ell'_\bu(u'))\big|^2\,d\mu(u)\cr
&=\int_{\bS^{2r-3}}\prod_{j=1}^{r-1}|\ell_j(e_j)|^2\times\sum_{j=0}^p
{(r-1)\,(p-j)!\,(r-2+j)!\over (p+r-1)!}
\,\big|s_j(\ell'_\bu(u'))\big|^2\,d\mu'(u'),\cr}
$$
since
$$
\int_0^1(1-t^2)^{p-j}t^{2r-3+2j}dt=
{1\over 2}\int_0^1(1-x)^{p-j}x^{r-2+j}dx=
{1\over 2}{(p-j)!\,(r-2+j)!\over (p+r-1)!}.
$$
Now, a standard inequality gives
$\max_j|\ell'_j(u')|\le 2\max_k|s_k(\ell'_\bu(u'))\big|^{1/k}$, thus
$$
\prod_{j=1}^p(|u_r|^2+|\ell'_j(u')|^2)\le
(|u_r|^2+M^2)^p=\sum_{k=0}^p{p\choose k}|u_r|^{2(p-k)}\,
|s_j(\ell'_\bu(u'))\big|^{k/j},
$$

(c) (d) Let $e_r=(0,0,\ldots,1)$. We write
$$
I(\ell_1,\ldots,\ell_r)=\int_{\bS^{2r-1}}
\prod_{j=1}^p|\ell_j(u')+\ell_j(e_r)u_r|^2\,d\mu(u).
$$
If we replace $u_r$ by $u_re^{i\theta}$ and average over $\theta$,
a use of Parseval's theorem for the trigonometric polynomial
$\prod(\ell_j(u')+\ell_j(e_r)u_re^{i\theta})$
implies
$$
I(\ell_1,\ldots,\ell_r)\ge\int_{\bS^{2r-1}}\prod_{j=1}^p|\ell_j(u')|^2\,d\mu(u)
={r-1\over p+r-1}\int_{\bS^{2r-3}}\prod_{j=1}^p|\ell_j'(u')|^2\,d\mu'(u')
$$
by (6.9), where $\ell'_j$ is the restriction of $\ell_j$ to the hyperplane
$e_r^\perp$, so that $|\ell'_j|^2=1-|\ell(e_r)|^2$. Using induction on
$r$ and a homogeneity argument, we find
$$
\int_{\bS^{2r-3}}\prod_{j=1}^p|\ell_j'(u')|^2\,d\mu'(u')\ge
m'_{r-1,p}\prod_{j=1}^p|\ell'_j|^2=
m'_{r-1,p}\prod_{j=1}^p(1-|\ell_j(e_r)|^2).
$$
This lower bound is in fact valid with any point $u\in\bS^{2r-1}$ replacing
$e_r$, thus
$$
I(\ell_1,\ldots,\ell_r)\ge {r-1\over p+r-1}\,m'_{r-1,p}
\prod_{j=1}^p(1-|\ell_j(u)|^2),\quad\forall u\in\bS^{2r-1}.
$$
By (6.9) again, we have $|u_r|\ge(1-\delta^2)^{1/2}$ on a set of measure
$$
\int_{u\in\bS^{2r-1},|u'|<\delta}d\mu(u)&=
\int_{v\in\bS^{2r-3}}\bigg(\int_0^\delta(2r-2)t^{2r-3}dt\bigg)d\mu'(v)
=\delta^{2r-2}.
$$
Therefore, for $\delta^2<p^{-1/2(r-1)}$, the set
$\bigcup\{|\ell_j(u)|\ge(1-\delta^2)^{1/2}\}$
does not cover $\bS^{2r-1}$, and we can find a point $u$ such that
$\prod(1-|\ell_j(u)|^2)\ge\delta^{2p}$. We get
$\max_{u\in\bS^{2r-1}}\prod(1-|\ell_j(u)|^2)\ge p^{-p/(r-1)}$ and
in this way, we infer inductively
$$
m'_{r,p}\ge {r-1\over p+r-1}\,p^{-p/(r-1)}\,m'_{r-1,p}.
$$
Since $m'_{1,p}=1$, we get
$$
m'_{r,p}\ge {p!\,(r-1)!\over (p+r-1)!}\,p^{-p(1+{1\over 2}+\cdots+{1\over r-1})}.
$$


For $p=1$, all results are trivial and
$m_{r,p}=m'_{r,p}={1\over r}$, so let us assume $p\ge 2$ and take 
$\ell_1,\ldots,\ell_p$ achieving the minimum. Let $\rho$ be 
the rank of $(\ell_1,\ldots,\ell_p)$. If $\rho=1$, we have
$I(\ell_1,\ldots,\ell_p)=m_{r,p}>m'_{r,p}$, contradiction,
thus $\rho\ge 2$. After taking a permutation, we can suppose that 
$\ell_1,\ldots,\ell_\rho$ are
linearly independent, and that $\ell_{\rho+1},\ldots,\ell_p$ are linear 
combinations of $\ell_1,\ldots,\ell_\rho$. By rotating coordinates, 
we can achieve that
$$
e_1=(1,0,\ldots,0)\in\bigcap_{2\le j\le\rho}\Ker(\ell_j)~~\hbox{and}~~
\ell_1(e_1)\ne 0.
$$
Then $\ell_j(e_1)=0$ for all $j\ge 2$. 
We~write 
$$
u'=(u_2,\ldots,u_r)\in\bC^{r-1},\quad
\ell_1(u)=\alpha_1 u_1+\ell'_1(u'),~\alpha_1\ne 0,~~\hbox{and }~~
\ell_j(u)=\ell'_j(u'),~j\ge 2.
$$
Then $|\ell_1|^2=|\alpha_1|^2+|\ell'_1|^2=1$ and $|\ell_j|^2=|\ell'_j|^2=1$ for
$j\ge 2$. Thanks to a symmetry argument $u_1\mapsto -u_1$, we find
$$
\eqalign{
m'_{p,r}=I(\ell_1,\ldots,\ell_p)
&=\int_{\bS^{2r-1}}
|\alpha_1u_1+\ell'_1(u')|^2\,|\ell'_2(u')|^2\ldots |\ell'_p(u')|^2\,d\mu(u)\cr
&=\int_{\bS^{2r-1}}
\big(|\alpha_1|^2|u_1|^2+|\ell'_1(u')|^2\big)\,
|\ell'_2(u')|^2\ldots |\ell'_p(u')|^2\,d\mu(u)\cr
&\ge |\alpha_1|^2m'_{r,p}+(1-|\alpha_1|^2)m'_{r,p}.\cr}
$$
The inequality cannot be strict, thus the minimum is achieved for both terms
$$
\int_{\bS^{2r-1}}
|u_1|^2\,|\ell'_2(u')|^2\ldots |\ell'_p(u')|^2\,d\mu(u),\quad
\int_{\bS^{2r-1}}
|\ell'_1(u')|^2\,|\ell'_2(u')|^2\ldots |\ell'_p(u')|^2\,d\mu(u).
$$


\noindent (b) We prove that by induction on $p$ that
$$
\prod_{j=1}^p(1+|a_j|^2)\le\sum_{k=0}^pc_{p,k}|s_k|^2
$$
where $c_{p,k}\in\bN$ and the $s_k$ are the elementary symmetric functions
in $a_1,\ldots,a_p$ (and by definition $s_0=1$). This is true for $p=1$ with
$c_{1,0}=c_{1,1}=1$. For $p=2$, we have
$$
1+|a_1|^2+|a_2|^2+|a_1|^2|a_2|^2\le
1+|a_1+a_2|^2+2|a_1|\,|a_2|+|a_1|^2|a_2|^2
\le 2+|a_1+a_2|^2+2|a_1|^2|a_2|^2,
$$
so one can take $c_{2,0}=2$, $c_{2,1}=1$, $c_{2,2}=2$. For $p=3$
$$
\eqalign{
&1+|a_1|^2+|a_2|^2+|a_3|^2+|a_1|^2|a_2|^2+|a_2|^2|a_3|^2+|a_1|^2|a_3|^2
+|a_1|^2|a_2|^2|a_3|^2\cr
&~~{}\le
1+|a_1+a_2+a_3|^2+2(|a_1||a_2|+|a_2||a_3|+|a_1||a_3|)\cr
&\kern5cm{}+|a_1|^2|a_2|^2+|a_2|^2|a_3|^2+|a_1|^2|a_3|^2
+|a_1|^2|a_2|^2|a_3|^2\cr
&~~{}\le
4+2|a_1+a_2+a_3|^2+2(|a_1|^2|a_2|^2+|a_2|^2|a_3|^2+|a_1|^2|a_3|^2)
+|a_1|^2|a_2|^2|a_3|^2\cr
&~~{}\le
4+2|a_1+a_2+a_3|^2+2|a_1a_2+a_2a_3+a_1a_3|^2
+4|a_1a_2a_3|\,|a_1+a_2+a_3|\cr
&\kern5cm{}+8|a_1|^2|a_2||a_3|+|a_1|^2|a_2|^2|a_3|^2\cr
&~~{}\le
4+2|a_1+a_2+a_3|^2+5|a_1a_2+a_2a_3+a_1a_3|^2
+5|a_1|^2|a_2|^2|a_3|^2,\cr}
$$
thus we can take $c_{3,0}=4$, $c_{3,1}=2$, $c_{3,2}=2$, $c_{3,3}=5$.
Assuming the result true for $p-1$, we multiply by $1+|a_p|^2$ and use
the fact that the new elementary symmetric functions are
$\tilde s_k=s_k+a_ps_{k-1}$. Thus
$$
(1+|a_p|^2)\sum_{k=0}^{p-1}c_{p-1,k}|s_k|^2=
(1+|a_p|^2)\sum_{k=0}^{p-1}c_{p-1,k}|\wt s_k|^2=
$$

\noindent
By expanding the product, we find
$$
\prod_{j=1}^p(1+|a_j|^2)=\sum_{k=0}^p\sigma_k,\quad\hbox{where}~~
\sigma_k:=\sum_{|I|=k}|a_I|^2.
$$
Now, we have
$$
\sigma_k=\sum_{|I|=k}|a_I|^2\le |s_k|^2+\bigg|\sum_{I\ne J}a_Ia_J\bigg|
$$
and if we observe that $a_Ia_J=a_{I\cup J}\,a_{I\cap J}$, we get
$\sum_{I\ne J}|a_I|\,|a_J|$ is bounded by
$$
\sum_{0\le j\le k-1}\bigg(
{p-j\choose 2k-2j}{2k-2j\choose k-j}
\sum_{|J|=j}|a_J|^2+{2k-j\choose j}{2k-2j\choose k-j}\sum_{|J|=2k-j}|a_J|^2
\bigg)
$$
and we find
$$
\sigma_k\le|s_k|^2+
\sum_{0\le j\le k-1}{p-j\choose 2k-2j}{2k-2j\choose k-j}\sigma_j
+{2k-j\choose j}{2k-2j\choose k-j}\sigma_{2k-j}
$$

\noindent (b) We consider the case $p=r$ and argue inductively.
For $r=1$, it is anyway clear that $m_{1,p}=m'_{1,p}=1$ for all $p$.
For $r=2$, (6.9) and $(6.9')$ imply
$$
\int_{\bS^3}|u_1|^4\,d\mu(u)={1\over 3},\quad
\int_{\bS^3}|u_1|^2\,|u_2|^2\,d\mu(u)={1\over 6}.
$$
For the general case $p=r=2$, we have to show that
$$
\int_{\bS^3}|u_1|^2\,|\alpha u_1+\beta u_2|^2\,d\mu(u)\ge{1\over 6}
$$
whenever $|\alpha|^2+|\beta|^2=1$, with equality if $\alpha=0$. However,
Parseval's formula yields
$$
\int_{\bS^3}|u_1|^2\,|\alpha u_1+\beta u_2|^2\,d\mu(u)=
\int_{\bS^3}|u_1|^2\,(|\alpha|^2|u_1|^2+|\beta|^2|u_2|^2)\,d\mu(u)
={1\over 3}|\alpha|^2+{1\over 6}|\beta|^2\ge {1\over 6},
$$
and we indeed have equality if and only if $\alpha$. We consider now
$r=3$. Then
$$
\eqalign{
&\int_{\bS^5}|u_1|^6\,d\mu(u)={1\over 10},\quad
\int_{\bS^5}|u_1|^4\,|u_2|^2\,d\mu(u)={1\over 30},\cr
&\int_{\bS^5}|u_1|^2\,|u_2|^2\,|u_3|^2\,d\mu(u)={1\over 60}.\cr}
$$
For the general case $p=r=3$, we have to show that
$$
\int_{\bS^5}|u_1|^2\,|\alpha u_1+\beta u_2|^2\,
|\gamma u_1+\delta u_2+\varepsilon u_3|^2\,d\mu(u)\ge{1\over 60}
$$
whenever $|\alpha|^2+|\beta|^2=1$ and
$|\gamma|^2+|\delta|^2+|\varepsilon|^2=1$, with equality if and only
if $\alpha=0$ and $\gamma=\delta=0$. However, this equal to
$$
\int_{\bS^5}|u_1|^2\,|\alpha u_1+\beta u_2|^2\,
(|\gamma u_1+\delta u_2|^2+|\varepsilon|^2|u_3|^2)\,d\mu(u)
$$
and the term $|\varepsilon|^2$ yields
$$
|\varepsilon|^2\int_{\bS^5}|u_1|^2\,|\alpha u_1+\beta u_2|^2\,
|u_3|^2\,d\mu(u)={1\over 10}
|\varepsilon|^2\int_{\bS^3}|u_1|^2\,|\alpha u_1+\beta u_2|^2\,d\mu(u)
\ge {|\varepsilon|^2\over 60}.
$$
(and equality implies $\alpha=0$, unless $\varepsilon=0$). We are left with
$$
\eqalign{
\int_{\bS^5}&|u_1|^2\,|\alpha u_1+\beta u_2|^2\,
|\gamma u_1+\delta u_2|^2\,d\mu(u)\cr
&=\int_{\bS^5}|u_1|^2\,\big|\alpha\gamma\,u_1^2+
(\alpha\delta+\beta\gamma)u_1u_2+\beta\delta\,u_2^2\big|^2\,d\mu(u)\cr
&=\int_{\bS^5}|u_1|^2\,\big(|\alpha|^2|\gamma|^2|u_1|^4+
|\alpha\delta+\beta\gamma|^2|u_1|^2|u_2|^2+|\beta|^2|\delta|^2|u_2|^4
\big)\,d\mu(u)\cr
&={1\over 10}|\alpha|^2|\gamma|^2
+{1\over 30}|\alpha\delta+\beta\gamma|^2+{1\over 30}|\beta|^2|\delta|^2\cr
&\ge{1\over 60}|\alpha|^2|\gamma|^2
+{1\over 60}\big(|\alpha|^2|\delta|^2+|\beta|^2|\gamma|^2-
2|\alpha|\,|\beta|\,|\gamma|\,|\delta|)+{1\over 60}|\beta|^2|\delta|^2\cr
&\quad{}+{5\over 60}|\alpha|^2|\gamma|^2
+{1\over 60}|\alpha\delta+\beta\gamma|^2+{1\over 60}|\beta|^2|\delta|^2
\cr
&={1\over 60}(|\alpha|^2+|\beta|^2)(|\gamma|^2+|\delta|^2)
+{4\over 60}|\alpha|^2|\gamma|^2
+{1\over 60}(|\alpha|\,|\gamma|-|\beta|\,|\delta|)^2\ge
{1\over 60}(|\gamma|^2+|\delta|^2).
\cr}
$$
Now equality implies $|\alpha|\,|\gamma|=|\beta|\,|\delta|=0$.
If $\gamma\ne 0$, then $\alpha=0$, $|\beta|=1$ and so $\delta=0$,
$|\gamma|^2=1-|\varepsilon|^2>0$. We are left with
${1\over 30}(1-|\varepsilon|^2)>{1\over 60}(1-|\varepsilon|^2)$
and equality cannot occur. Hence $\gamma=0$ and
$|\delta|^2=1-|\varepsilon|^2$. If $\delta\ne 0$,
then $\beta=0$, $|\alpha|=1$; we are
left again with ${1\over 30}(1-|\varepsilon|^2)$, and equality cannot occur.
Therefore $\gamma=\delta=0$, $|\varepsilon|=1$ and $\alpha=0$.

\noindent
For $r=4$, we have
$$
\eqalign{
&\int_{\bS^7}|u_1|^8\,d\mu(u)={1\over 35},\quad
\int_{\bS^7}|u_1|^6\,|u_2|^2\,d\mu(u)={1\over 140},\cr
&\int_{\bS^7}|u_1|^4\,|u_2|^2\,|u_3|^2\,d\mu(u)={1\over 420},\quad
\int_{\bS^7}|u_1|^2\,|u_2|^2\,|u_3|^2\,|u_4|^2\,d\mu(u)={1\over 840}.\quad
\cr}
$$
For the general case $p=r=4$, we have to compute
$$
\eqalign{
\int_{\bS^7}&|u_1|^2\,|\alpha u_1+\beta u_2|^2\,
|\gamma u_1+\delta u_2+\varepsilon u_3|^2\,
|\zeta u_1+\eta u_2+\theta u_3+\iota u_4|^2\,
d\mu(u)\cr
&=\int_{\bS^7}|u_1|^2\,|\alpha u_1+\beta u_2|^2\,
|\gamma u_1+\delta u_2+\varepsilon u_3|^2\,
\big(|\zeta u_1+\eta u_2+\theta u_3|^2+|\iota|^2|u_4|^2\big)\,
d\mu(u)\cr}
$$
and the term in $|\iota|^2|u_4|^2$ has the appropriate lower bound
${1\over 840}|\iota|^2$, with the ad hoc equality case if $\iota\ne 0$.
We are left with the case $\iota=0$, leading to
$$
\eqalign{\kern-45pt
&\int_{\bS^7}|u_1|^2\,|\alpha u_1+\beta u_2|^2\,\Big|
(\gamma u_1+\delta u_2)(\zeta u_1+\eta u_2)+
\big(\theta(\gamma u_1+\delta u_2)+\varepsilon(\zeta u_1+\eta u_2)\big)u_3+
\varepsilon\theta\,u_3^2\Big|^2\,d\mu(u)\cr
\kern-45pt&=\int_{\bS^7}|u_1|^2\,|\alpha u_1+\beta u_2|^2\,\Big(|
\gamma u_1+\delta u_2|^2|\zeta u_1+\eta u_2|^2+
\big|\theta(\gamma u_1+\delta u_2)+\varepsilon(\zeta u_1+\eta u_2)\big|^2\,
|u_3|^2+
|\varepsilon|^2\,|\theta|^2\,|u_3|^4\Big)\,d\mu(u),\cr}
$$
and, in particular, we have to consider
$$
\int_{\bS^7}|u_1|^2\,|\alpha u_1+\beta u_2|^2\,
|\gamma u_1+\delta u_2|^2|\zeta u_1+\eta u_2|^2\,d\mu(u).
$$

\noindent (b) We first consider the case $p\le r$. Assume that
$\ell_1,\ldots,\ell_p$ have been chosen such that $I(\ell_1,\ldots,\ell_p)$
is a local minimum. We claim that the $\ell_1,\ldots,\ell_p$ are orthogonal.
if $\ell_1,\ldots,\ell_p$ are linearly dependent, we take $\lambda_j\in
\ell_j^\perp$, $|\lambda_j|=1$ and put
$$
\ell_{j,t,\theta}=
(1+t^2|\alpha_j|^2)^{-1/2}(\ell_j+t\,e^{i\theta}\alpha_j\lambda_j),\quad
0<t\ll 1,~~\theta\in[0,2\pi].
$$
Then $|\ell_{j,t,\theta}|=1$ and
$$
\prod_{j=1}^p|\ell_{j,t,\theta}|^2=\prod_{j=1}^p(1+t^2|\alpha_j|^2)^{-1}\,
\bigg|\ell_1\ldots\ell_p\bigg
(1+\sum_{k=1}^pt^ke^{ki\theta}s_k\bigg)\bigg|^2
$$
where the $s_k$ are the elementary symmetric functions of the
ratios $\alpha_j\lambda_j/\ell_j$. Since $(\ell_1,\ldots,\ell_p)$ must be
a critical point, we have
$$
\int_{\bS^{2r-1}}|\ell_1|^2\ldots\,|\ell_p|^2\,
s_1((\alpha\lambda/\ell)_\bu)\,d\mu(u)=0
$$
for all possible choices of the constants, i.e.
$$
\int_{\bS^{2r-1}}|\ell_1|^2\ldots\,|\ell_p|^2\,{\lambda_j\over\ell_j}
\,d\mu(u)=0,\quad\forall j,~\forall\lambda_j.
$$
The relevant term in the expansion of the product is the term $k=2$,namely
$$
t^2e^{2i\theta}\,\ell_1\ldots\ell_p\sum_{j<k}{\alpha_j\alpha_k\lambda_j\lambda_k
\over\ell_j\ell_k}
$$
and the Hessian form of $I(\ell_1,\ldots,\ell_p)$ along the complex line
$(te^{i\theta}\alpha_j\lambda_j)$ is given by
$$
t\mapsto
\int_{\bS^{2r-1}}\kern-4pt
|\ell_1|^2\ldots\,|\ell_p|^2\,\bigg(1-\sum_jt^2|\alpha_j|^2
+2\Re\bigg(t^2e^{2i\theta}\sum_{j<k}{\alpha_j\alpha_k\lambda_j\lambda_k
\over\ell_j\ell_k}\bigg)+t^2\bigg|\sum_j{\alpha_j\lambda_j\over\ell_j}\bigg|^2
\bigg)d\mu(u).
$$




If the rank of $(\ell_1,\ldots,\ell_p)$ is${}<r$,
we may assume after rotating coordinates
that $\ell_1,\ldots,\ell_p$ depend only on $u'=(u_1,\ldots,u_{r-1})$. Then
formula (5.26) shows that the equality case is reached in dimension~$r-1$;
by induction, we conclude that $p\le r-1$ and that the $\ell_j$ are
orthogonal. We are left with the case where the rank of $\ell_1,\ldots,\ell_p$
is equal to $r$. By permuting the $\ell_j$'s, we can assume that
$\ell_1,\ldots,\ell_j$ are linear independent, and that
$$
\ell_1
$$

By our calculation of the integral
of $|u_1|^2\dlots|u_p|^2$, the inequality is an equality whenener
$\ell_1,\ldots,\ell_p$ are orthogonal. Now, assume that there is
one linear form $\ell_j=\langle\bu,a_j\rangle$ such that $

Let us consider $\smash{\max_{u\in\bS^{2r-1}}\prod_{j=1}^p|\ell_j(u)|^2}$.
The maximum is attained at a certain point of $\bS^{2r-1}$, and by rotating
coordinates, we can assume that it occurs at $e_r=(0,0,\ldots,1)$. We
write $u=u'+u_re_r$ and $\ell_j(u)=\ell_j(u')+\ell_j(e_r)u_r$.
If we replace $u_r$ by $u_re^{i\theta}$ and average over $\theta\in[0,2\pi]$,
a use of Parseval's theorem for the trigonometric
polynomial $\prod_j(\ell_j(u')+\ell_j(e_r)u_re^{i\theta})$
implies
$$
I(\ell_1,\ldots,\ell_p)\ge
\prod_{j=1}^p|\ell_j(e_r)|^2\int_{\bS^{2r-1}}|u_r|^{2p}\,d\mu(u)=
m_{r,p}\,\max_{u\in\bS^{2r-1}}\prod_{j=1}^p|\ell_j(u)|^2.
$$
However, (6.9) shows that we have $|u_r|\le\delta$ on a set of measure
$$
\int_{u\in\bS^{2r-1},|u_r|<\delta}d\mu(u)=
\int_{u'\in\bS^{2r-3}}\bigg(\int_{(1-\delta^2)^{1/2}}^1
(2r-2)t^{2r-3}dt\bigg)d\mu'(u')=1-(1-\delta^2)^{r-1},
$$
thus this is less than $1/p$ for $\delta^2<1-(1-{1\over p})^{1/(r-1)}$.
In this circumstance, the set $\bigcup\{|\ell_j(u)|\le\delta\}$
does not cover $\bS^{2r-1}$, and we can find a point $u$ such that
$\prod|\ell_j(u)|^2\ge\delta^{2p}$. Therefore
$\max_{u\in\bS^{2r-1}}\prod|\ell_j(u)|^2\ge(1-(1-{1\over p})^{1/(r-1)})^p\ge
({1\over p(r-1)})^p$ and
$$
I(\ell_1,\ldots,\ell_p)\ge {p!\,(r-1)!\over (p+r-1)!}\,
\Big(1-\Big(1-{1\over p}\Big)^{1/(r-1)}\Big)^p.
$$

\noindent (b) For $r=2$, we have to show that
for all linear forms $\ell_1,\ldots,\ell_p\in(\bC^2)^*$, the inequality
$$
\int_{\bS^3}\prod_{j=1}^p|\ell_j(u)|^2\,d\mu(u)\ge
{1\over (p+1)!}\prod_{j=1}^p|\ell_j|^2\leqno(5.27)
$$
holds. It will be useful to observe that (5.27) is invariant under the action 
of $U(2)$ on $\bC^2$. By homogeneity, we can assume that $\ell_j(z)=u_1+a_ju_2$,
$1\le j\le p$. Then formula (5.27) reads
$$
\int_{\bS^3}\bigg|\sum_{k=0}^p s_ku_1^{p-k}u_2^k\bigg|^2\,d\mu(u)\ge
{1\over (p+1)!}\prod_{j=1}^p(1+|a_j|^2),\leqno(5.27')
$$
where $s_k$ is the $k$-th elementary symmetric function in $a_1,\ldots,a_p$,
and $s_0=1$. If we apply rotations $(u_1,u_2)\mapsto (e^{i\theta_1}u_1,
e^{i\theta_2}u_2)$ and use Parseval's formula, this is equivalent to
$$
\int_{\bS^3}\sum_{k=0}^p|s_k|^2|u_1|^{2p-2k}|u_2|^{2k}\,d\mu(u)\ge
{1\over (p+1)!}\prod_{j=1}^p(1+|a_j|^2).\leqno(5.27'')
$$
Now, (5.25) implies
$$
\int_{\bS^3}|u_1|^{2p-2k}|u_2|^{2k}\,d\mu(u)
=2\int_0^1t^{2p-2k}(1-t)^{2k}\,t\,dt
=\int_0^1t^{p-k}(1-t)^k\,dt={k!\,(p-k)!\over(p+1)!},
$$
and we see that $(5.27)$, $(5.27')$ or $(5.27'')$ are equivalent to 
the ``elementary'' inequality
$$
\prod_{j=1}^p(1+|a_j|^2)\le \sum_{k=0}^p k!\,(p-k)!\,|s_k|^2,
\leqno(5.28)
$$
relating a polynomial $X^p-s_1X^{p-1}+\cdots+(-1)^ps_p$ and its complex
roots~$a_j$. Now, we prove (5.28) by induction on $p$, the case $p=1$
being trivial. Assume that (5.28) holds for $p-1$, with some $p\ge 2$.
Then, by the $U(2)$ invariance, we can assume that $\ell_p(u)=u_1$,
i.e.\ $a_p=0$. The induction hypothesis for $p-1$ implies
$$
\prod_{j=1}^{p-1}(1+|a_j|^2)\le \sum_{k=0}^{p-1} k!\,(p-1-k)!\,|s_k|^2,
$$
and it is clear that $(5.28)$ then holds for $p$, since $a_p=0$ and the 
elementary symmetric functions are kept unchanged. It should be observed 
that (5.28) is not optimal symptotically when $p\to+\infty\,$; in fact,
Landau's inequality [Land05] gives
$\prod\max(1,|a_j|)\le(\sum|s_k|^2)^{1/2}$, from which one
can easily derive that $\prod(1+|a_j|^2\le 2^p\sum|s_k|^2$, which improves
(5.28) as soon as $p\ge 7$ (observe that $2^7=128$ and $3!\,4!=144$). However,
(5.28) is sharper for $p\le 3$.

\noindent (b) We first assume $p\le r$ and show that the minimum is reached
exactly when the forms $\ell_j$ are orthogonal. For this, we argue by induction
on $r$, the statement being valid (and void!) for $r=1$. If $p\le r-1$,
we can find orthonormal coordinates
$u=(u',u_r)$, $u'=(u_1,\ldots,u_{r-1})$, such that our forms depend only on
$u'$, and we use (5.26) to conclude  that $\ell_1,\ldots,\ell_p$ must be
orthogonal. Now, assume that $p=r$. If $\ell_1,\ldots,\ell_r$ are linearly
dependent, we choose coordinates so that $e_r=(0,\ldots,0,1)\in\bigcap\Ker
\ell_j$ and coefficients $\alpha_j$ such that $\sum\alpha_j\ell_j=0$,
where not all $\alpha_j$ vanish. We write $u=u'+u_re_r$ and put
$$
\wt\ell_j(u)=(1-t^2|\alpha_j|^2)^{1/2}\ell_j(u')+t\overline\alpha_ju_r,\quad t>0.
$$
Then $|\wt\ell_j|=1$ and
$$
\prod_{j=1}^p|\wt\ell_j(u)|^2=
\prod_{j=1}^p(1-t^2|\alpha_j|^2)\,|\ell_j(u')|^2
+\sum_{0\le j,k\le p,\,(j,k)\ne(0,0)}t^{j+k}u_r^j\,\overline u_r^k\,
s_{j,k}(t,u',\overline u')
$$
where $s_{j,k}$ is a polynomial. The leading terms of interest are
$$
s_{1,0}(t,u',\overline u')=\prod\ell_j
$$

\noindent (b) We have
$$
dI(\ell_1,\ldots,\ell_p)(h_1,\ldots,h_p)=
2\sum_{j=1}^p\int_{\bS^{2r-1}}\prod_{k\ne j}|\ell_k(u)|^2\,
\Re\big(\ell_j(u)\overline{h_j(u)}\,\big)\,d\mu(u)
$$
Now, we prove by
induction on $r$ that for arbitrary $l_j\in(\bC^r)^*$ we have
$$
I(\ell_1,\ldots,\ell_p)\ge {(r-1)!\over(p+r-1)!}\prod_{j=1}^p|\ell_j|^2.
\leqno(5.29)
$$
The result is clear for $r=1$, and we have just proved it for $r=2$.
Without loss of generality, thanks to the $U(r)$ invariance,
we can assume that $e_r=(0,\ldots,0,1)\in
\bigcap_{p-r+2\le j\le p}\Ker\ell_j$. We reorder the linear forms
in such a way that $\ell_j(e_r)\ne 0$ for $1\le j\le q$ and
$\ell_j(e_r)=0$ for $q+1\le j\le p$, with $q\le p-r+1$. We write
$u=(u',u_r)\in\bC^{r-1}\times\bC$ and $\ell_j(u)=\ell'_j(u')+\ell(e_r)u_r$
where $\ell'_j$ is the restriction of $\ell_j$ to $\bC^{r-1}\times\{0\}\simeq
\bC^{r-1}$. By homogeneity, we can in fact assume that $\ell(e_j)=1$ for
$1\le j\le q$. Then Parseval's formula implies
$$
\eqalign{
I(\ell_1,\ldots,\ell_p)
&=\int_{\bS^{2r-1}}\prod_{j=1}^q|\ell'_j(u')+u_r|^2\,
\prod_{j=q+1}^p|\ell_j(u')|^2\,d\mu(u)\cr
&=\int_{\bS^{2r-1}}\bigg|
\sum_{k=0}^qs_k(\ell'(u'))\,u_r^{q-k}\bigg|^2
\prod_{j=q+1}^p|\ell'_j(u')|^2\,d\mu(u)\cr
&\ge\int_{\bS^{2r-1}}{1\over (q+1)!}\prod_{j=1}^q
\big(\big|\ell'_j(u')\big|^2+|u_r|^2\big)
\prod_{j=q+1}^p|\ell'_j(u')|^2\,d\mu(u)\cr}
$$


$$
\eqalign{
I(\ell_1,\ldots,\ell_p)
&=\int_{\bS^{2r-1}}\prod_{j=1}^q|\ell'_j(u')+u_r|^2\,
\prod_{j=q+1}^p|\ell_j(u')|^2\,d\mu(u)\cr
&=\int_{\bS^{2r-1}}\bigg|
\sum_{k=0}^qs_k(\ell'(u'))\,u_r^{q-k}\bigg|^2
\prod_{j=q+1}^p|\ell'_j(u')|^2\,d\mu(u)\cr
&=\int_{\bS^{2r-1}}
\sum_{k=0}^q\big|s_k(\ell'(u'))\big|^2\,|u_r|^{2(q-k)}\,
\prod_{j=q+1}^p|\ell'_j(u')|^2\,d\mu(u)\cr
&=\int_{\bS^{2r-3}}
\sum_{k=0}^q{(r-1)\,(k+r-1)!(q-k)!\over(q+r-1)!}\,
\big|s_k(\ell'(u'))\big|^2\,
\prod_{j=q+1}^p|\ell'_j(u')|^2\,d\mu'(u'),\cr}
$$
where we have used (5.25) and the fact that
$$
(2r-2)\int_0^1t^{2k}(1-t^2)^{q-k}t^{2r-3}\,dt
={(r-1)\,(k+r-1)!\,(q-k)!\over(q+r-1)!}.
$$
Since $(k+r-1)!\ge k!\,(r-1)!$, our inequality (5.28) implies
$$
\eqalign{
I(\ell_1,\ldots,\ell_p)
&\ge(r-1)\,(r-1)!\int_{\bS^{2r-3}}\sum_{k=0}^q{k!(q-k)!\over(q+r-1)!}
\big|s_k(\ell'(u'))\big|^2\,
\prod_{j=q+1}^p|\ell'_j(u')|^2\,d\mu'(u'),\cr
&\ge{(r-1)\,(r-1)!\over (q+r-1)!}\int_{\bS^{2r-3}}
\prod_{j=1}^q(1+|\ell'_j(u')|^2)\,d\mu(u')
\prod_{j=q+1}^p|\ell'_j(u')|^2\,d\mu'(u').\cr}
$$
Now, we expand
$$
\prod_{j=1}^q(1+|\ell'_j(u')|^2)\prod_{j=q+1}^p|\ell'_j(u')|^2=
\sum_{k=0}^q~\sum_{|I|=k}~\prod_{j\in J}|\ell'_j(u')|^2
\prod_{j=q+1}^p|\ell'_j(u')|^2.
$$
and use the induction hypothesis for $r'=r-1$ and $p'=p-q+k$. This gives
$$
I(\ell_1,\ldots,\ell_p)\ge
{(r-1)\,(r-1)!\over (q+r-1)!}\sum_{k=0}^q~{(r-2)!\over (p-q+k+r-2)!}\sum_{|I|=k}
~\prod_{j\in J}|\ell'_j(u')|^2
$$

(c) We prove the lower bound by induction on $r$. For $r=1$,
we have $\ell_j(u)=\alpha_ju_1$ with $\alpha_j=1$ and it is obvious that
one can take $m'_{1,p}=1$. Let us now assume $r\ge 2$.
We write
$$
\eqalign{
I(\ell_1,\ldots,\ell_p)
&=\int_{\bS^{2r-1}}\prod_{j=1}^p|\ell_j(u')+\ell_j(e_r)u_r|^2\,d\mu(u)\cr
&=\int_{\bS^{2r-1}}\prod_{j=1}^p|\ell_j(e_r)|^2~
\Bigg|\sum_{j=0}^pu_r^{p-j}s_j(\ell'_\bu(u'))\Bigg|^2d\mu(u)\cr}
$$
where $\ell'_j(u')=\ell_j(u')/\ell_j(e_r)$ and the $s_j(\bu)$ denote the 
elementary symmetric functions. In the last equality, we make a change
of variable $u_r\mapsto e^{i\theta}u_r$ and take the average on
$\theta\in[0,2\pi]$ by means of Parseval's equality. By (6.9) we get
$$
\eqalign{
I(\ell_1,\ldots,\ell_p)
&=\int_{\bS^{2r-1}}\prod_{j=1}^{r-1}|\ell_j(e_r)|^2\times\sum_{j=0}^p
|u_r|^{2(p-j)}\,\big|s_j(\ell'_\bu(u'))\big|^2\,d\mu(u)\cr
&=\int_{\bS^{2r-3}}\prod_{j=1}^{r-1}|\ell_j(e_r)|^2\times\sum_{j=0}^p
{(r-1)\,(p-j)!\,(r-2+j)!\over (p+r-1)!}
\,\big|s_j(\ell'_\bu(u'))\big|^2\,d\mu'(u'),\cr}
$$
since
$$
\int_0^1(1-t^2)^{p-j}t^{2r-3+2j}dt=
{1\over 2}\int_0^1(1-x)^{p-j}x^{r-2+j}dx=
{1\over 2}{(p-j)!\,(r-2+j)!\over (p+r-1)!}.
$$
By induction on $r$, we also have
$$
\eqalign{
\int_{\bS^{2r-1}}|\ell'_{m_1}(u')|^2\ldots\,&|\ell'_{m_j}(u')|^2\,|u_r|^{2(p-j)}\,
d\mu(u)\cr
&={(r-1)\,(p-j)!\,(r-2+j)!\over (p+r-1)!}
\int_{\bS^{2r-3}}|\ell'_{m_1}(u')|^2\ldots|\ell'_{m_j}(u')|^2\,d\mu(u')\cr
&\ge{(r-1)\,(p-j)!\,(r-2+j)!\over (p+r-1)!}\,
{(r-2)!\over (r-2+j)!}\prod_{i=1}^j|\ell'_{m_i}|^2\cr
&={(p-j)!\,(r-1)!\over (p+r-1)!}\,
\prod_{i=1}^j{1-|\ell_{m_i}(e_r)|^2\over|\ell_{m_i}(e_r)|^2},\cr}
$$
and we want to show that
$$
\eqalign{
&\int_{\bS^{2r-3}}\prod_{j=1}^{r-1}|\ell_j(e_r)|^2\times\sum_{j=0}^p
{(r-1)\,(p-j)!\,(r-2+j)!\over (p+r-1)!}
\,\big|s_j(\ell'_\bu(u'))\big|^2\,d\mu'(u'),\cr
&\ge {r-1!\over (p+r-1)!}=
{r-1!\over (p+r-1)!}\prod_{j=1}^p|\ell_j(e_r)|^2
\prod_{j=1}^p\Big(1+{1-|\ell_j(e_r)|^2\over|\ell_j(e_r)|^2}\Big)
\cr
&={r-1!\over (p+r-1)!}\prod_{j=1}^p|\ell_j(e_r)|^2
\sum_{j=0}^p~\sum_{1\le m_1<\ldots<m_j\le p}~
\prod_{i=1}^j{1-|\ell_{m_i}(e_r)|^2\over|\ell_{m_i}(e_r)|^2},
\cr}
$$
so, it suffices to show that
$$
\eqalign{
\sum_{j=0}^p&
(p-j)!\,(r-2+j)!
\,\big|s_j(\ell'_\bu(u'))\big|^2\cr
&\ge\sum_{j=0}^p~\sum_{1\le m_1<\ldots<m_j\le p}~
(r-2+j)!\,|\ell'_{m_1}(u')|^2\ldots|\ell'_{m_j}(u')|^2,
\cr}
$$
or, we can show instead that
$$
\eqalign{
\sum_{j=0}^p&
|u_r|^{2(p-j)}
\,\big|s_j(\ell'_\bu(u'))\big|^2\cr
&\ge\sum_{j=0}^p~\sum_{1\le m_1<\ldots<m_j\le p}~
{|u_r|^{2(p-j)}\over (p-j)!}\,|\ell'_{m_1}(u')|^2\ldots|\ell'_{m_j}(u')|^2.
\cr}
$$

Now, a standard inequality gives
$\max_j|\ell'_j(u')|\le 2\max_k|s_k(\ell'_\bu(u'))\big|^{1/k}$, thus
$$
\prod_{j=1}^p(|u_r|^2+|\ell'_j(u')|^2)\le
(|u_r|^2+M^2)^p=\sum_{k=0}^p{p\choose k}|u_r|^{2(p-k)}\,
|s_j(\ell'_\bu(u'))\big|^{k/j},
$$

\noindent (b) We consider the case $p=r$ and argue inductively.
For $r=1$, it is anyway clear that $m_{1,p}=m'_{1,p}=1$ for all $p$.
For $r=2$, (6.9) and $(6.9')$ imply
$$
\int_{\bS^3}|u_1|^4\,d\mu(u)={1\over 3},\quad
\int_{\bS^3}|u_1|^2\,|u_2|^2\,d\mu(u)={1\over 6}.
$$
For the general case $p=r=2$, we have to show that
$$
\int_{\bS^3}|u_1|^2\,|\alpha u_1+\beta u_2|^2\,d\mu(u)\ge{1\over 6}
$$
whenever $|\alpha|^2+|\beta|^2=1$, with equality if $\alpha=0$. However,
Parseval's formula yields
$$
\int_{\bS^3}|u_1|^2\,|\alpha u_1+\beta u_2|^2\,d\mu(u)=
\int_{\bS^3}|u_1|^2\,(|\alpha|^2|u_1|^2+|\beta|^2|u_2|^2)\,d\mu(u)
={1\over 3}|\alpha|^2+{1\over 6}|\beta|^2\ge {1\over 6},
$$
and we indeed have equality if and only if $\alpha$. We consider now
$r=3$. Then
$$
\eqalign{
&\int_{\bS^5}|u_1|^6\,d\mu(u)={1\over 10},\quad
\int_{\bS^5}|u_1|^4\,|u_2|^2\,d\mu(u)={1\over 30},\cr
&\int_{\bS^5}|u_1|^2\,|u_2|^2\,|u_3|^2\,d\mu(u)={1\over 60}.\cr}
$$
For the general case $p=r=3$, we have to show that
$$
\int_{\bS^5}|u_1|^2\,|\alpha u_1+\beta u_2|^2\,
|\gamma u_1+\delta u_2+\varepsilon u_3|^2\,d\mu(u)\ge{1\over 60}
$$
whenever $|\alpha|^2+|\beta|^2=1$ and
$|\gamma|^2+|\delta|^2+|\varepsilon|^2=1$, with equality if and only
if $\alpha=0$ and $\gamma=\delta=0$. However, this equal to
$$
\int_{\bS^5}|u_1|^2\,|\alpha u_1+\beta u_2|^2\,
(|\gamma u_1+\delta u_2|^2+|\varepsilon|^2|u_3|^2)\,d\mu(u)
$$
and the term $|\varepsilon|^2$ yields
$$
|\varepsilon|^2\int_{\bS^5}|u_1|^2\,|\alpha u_1+\beta u_2|^2\,
|u_3|^2\,d\mu(u)={1\over 10}
|\varepsilon|^2\int_{\bS^3}|u_1|^2\,|\alpha u_1+\beta u_2|^2\,d\mu(u)
\ge {|\varepsilon|^2\over 60}.
$$
(and equality implies $\alpha=0$, unless $\varepsilon=0$). We are left with
$$
\eqalign{
\int_{\bS^5}&|u_1|^2\,|\alpha u_1+\beta u_2|^2\,
|\gamma u_1+\delta u_2|^2\,d\mu(u)\cr
&=\int_{\bS^5}|u_1|^2\,\big|\alpha\gamma\,u_1^2+
(\alpha\delta+\beta\gamma)u_1u_2+\beta\delta\,u_2^2\big|^2\,d\mu(u)\cr
&=\int_{\bS^5}|u_1|^2\,\big(|\alpha|^2|\gamma|^2|u_1|^4+
|\alpha\delta+\beta\gamma|^2|u_1|^2|u_2|^2+|\beta|^2|\delta|^2|u_2|^4
\big)\,d\mu(u)\cr
&={1\over 10}|\alpha|^2|\gamma|^2
+{1\over 30}|\alpha\delta+\beta\gamma|^2+{1\over 30}|\beta|^2|\delta|^2\cr
&\ge{1\over 60}|\alpha|^2|\gamma|^2
+{1\over 60}\big(|\alpha|^2|\delta|^2+|\beta|^2|\gamma|^2-
2|\alpha|\,|\beta|\,|\gamma|\,|\delta|)+{1\over 60}|\beta|^2|\delta|^2\cr
&\quad{}+{5\over 60}|\alpha|^2|\gamma|^2
+{1\over 60}|\alpha\delta+\beta\gamma|^2+{1\over 60}|\beta|^2|\delta|^2
\cr
&={1\over 60}(|\alpha|^2+|\beta|^2)(|\gamma|^2+|\delta|^2)
+{4\over 60}|\alpha|^2|\gamma|^2
+{1\over 60}(|\alpha|\,|\gamma|-|\beta|\,|\delta|)^2\ge
{1\over 60}(|\gamma|^2+|\delta|^2).
\cr}
$$
Now equality implies $|\alpha|\,|\gamma|=|\beta|\,|\delta|=0$.
If $\gamma\ne 0$, then $\alpha=0$, $|\beta|=1$ and so $\delta=0$,
$|\gamma|^2=1-|\varepsilon|^2>0$. We are left with
${1\over 30}(1-|\varepsilon|^2)>{1\over 60}(1-|\varepsilon|^2)$
and equality cannot occur. Hence $\gamma=0$ and
$|\delta|^2=1-|\varepsilon|^2$. If $\delta\ne 0$,
then $\beta=0$, $|\alpha|=1$; we are
left again with ${1\over 30}(1-|\varepsilon|^2)$, and equality cannot occur.
Therefore $\gamma=\delta=0$, $|\varepsilon|=1$ and $\alpha=0$.

\noindent
For $r=4$, we have
$$
\eqalign{
&\int_{\bS^7}|u_1|^8\,d\mu(u)={1\over 35},\quad
\int_{\bS^7}|u_1|^6\,|u_2|^2\,d\mu(u)={1\over 140},\cr
&\int_{\bS^7}|u_1|^4\,|u_2|^2\,|u_3|^2\,d\mu(u)={1\over 420},\quad
\int_{\bS^7}|u_1|^2\,|u_2|^2\,|u_3|^2\,|u_4|^2\,d\mu(u)={1\over 840}.\quad
\cr}
$$
For the general case $p=r=4$, we have to compute
$$
\eqalign{
\int_{\bS^7}&|u_1|^2\,|\alpha u_1+\beta u_2|^2\,
|\gamma u_1+\delta u_2+\varepsilon u_3|^2\,
|\zeta u_1+\eta u_2+\theta u_3+\iota u_4|^2\,
d\mu(u)\cr
&=\int_{\bS^7}|u_1|^2\,|\alpha u_1+\beta u_2|^2\,
|\gamma u_1+\delta u_2+\varepsilon u_3|^2\,
\big(|\zeta u_1+\eta u_2+\theta u_3|^2+|\iota|^2|u_4|^2\big)\,
d\mu(u)\cr}
$$
and the term in $|\iota|^2|u_4|^2$ has the appropriate lower bound
${1\over 840}|\iota|^2$, with the ad hoc equality case if $\iota\ne 0$.
We are left with the case $\iota=0$, leading to
$$
\eqalign{\kern-45pt
&\int_{\bS^7}|u_1|^2\,|\alpha u_1+\beta u_2|^2\,\Big|
(\gamma u_1+\delta u_2)(\zeta u_1+\eta u_2)+
\big(\theta(\gamma u_1+\delta u_2)+\varepsilon(\zeta u_1+\eta u_2)\big)u_3+
\varepsilon\theta\,u_3^2\Big|^2\,d\mu(u)\cr
\kern-45pt&=\int_{\bS^7}|u_1|^2\,|\alpha u_1+\beta u_2|^2\,\Big(|
\gamma u_1+\delta u_2|^2|\zeta u_1+\eta u_2|^2+
\big|\theta(\gamma u_1+\delta u_2)+\varepsilon(\zeta u_1+\eta u_2)\big|^2\,
|u_3|^2+
|\varepsilon|^2\,|\theta|^2\,|u_3|^4\Big)\,d\mu(u),\cr}
$$
and, in particular, we have to consider
$$
\int_{\bS^7}|u_1|^2\,|\alpha u_1+\beta u_2|^2\,
|\gamma u_1+\delta u_2|^2|\zeta u_1+\eta u_2|^2\,d\mu(u).
$$

$$
{1\over 360}(1+|a|^2)(1+|b|^2)(1+|d|^2)
+{1\over 180}(2+|a|^2)|c|^2|d|^2
+{1\over 360}(1+|a|^2)(|e+c|^2+|eb+cd|^2)
$$

$$
\ge {1\over 360}(1+|a|^2)\Big((1+|b|^2)(1+|d|^2)
+2|c|^2|e|^2+|e+c|^2+|eb+cd|^2\Big)
$$

$$
\ge {1\over 360}(1+|a|^2)(1+|b|^2+|c|^2)(1+|d|^2+|e|^2)~??
$$