Theorem. Let $V$ a centered Gaussian variable in $R^n$ with covariance matrix $\Sigma.$ Let
\begin{equation}\label{J}J_n=\frac{1}{2\sqrt{\pi}}\int_0^{\infty}\left(1-\left(1+\frac{2s}{n}\right)^{-n/2}\right)\frac{ds}{s^{3/2}}.\end{equation} Then for all $\Sigma$
\begin{equation}\label{BASE}\frac{E(\|V\|)}{\sqrt{E(\|V\|^2)}}\geq J_n \ \ (A)\end{equation} and equality holds if and only if $\Sigma$ is proportional to the identity matrix $I_n.$ Furthermore
$J_n$ increases to 1. In particular $\frac{E(\|V\|)}{\sqrt{E(\|V\|^2)}}\geq J_1=\frac{\sqrt{2}}{\sqrt{\pi}}$ which implies the Kintchine-Kahane inequality $$\mathrm{Var}(\|V\|)\leq \left(\frac{\pi}{2}-1\right)[E(\|V\|)]^2\leq [E(\|V\|)]^2.$$
Proof.
Write $\Sigma=U^TDU$ where $U$ is orthogonal
and $D=\mathrm{diag}(a_1, \ldots ,a_n)$ Therefore if $Z\sim N(0,I_n)$ we have $V\sim \Sigma^{1/2}Z\sim D^{1/2}Z$ since $Z\sim UZ.$
and therefore \begin{equation}\label{NORME}\|V\|^2\sim a_1Z_1^2+\cdots+a_nZ_n^2\ \ (B)\end{equation} where $Z_1,\ldots,Z_n$ are independent and $N(0,1).$ This implies that $E(\|V\|^2)=a_1+\cdots+a_n=\mathrm{trace}\, \Sigma.$
Now we embark for a minoration of $E(\|V\|)=E(( a_1Z_1^2+\cdots+a_nZ_n^2)^{1/2})$ For this we use the following integral representation of the square root of the positive number $x:$
\begin{equation}\label{INTEGRAL}\sqrt{x}=\frac{1}{2\sqrt{\pi}}\int_0^{\infty}\frac{1-e^{-sx}}{s^{3/2}}ds \ \ (C)\end{equation} For proving it we observe that (C) is correct for $x=0$ and that the derivatives of both sides coincide. As a consequence of (C) we have
\begin{eqnarray*}E(\|V\|)&=&\frac{1}{2\sqrt{\pi}}\int_0^{\infty}(1-E(e^{-s(a_1Z_1^2+\cdots+a_nZ_n^2)})\frac{ds}{s^{3/2}}\\&=&\frac{1}{2\sqrt{\pi}}\int_0^{\infty}\left(1-\prod_{i=1}^nE(e^{-sa_iZ_i^2}\right)\frac{ds}{s^{3/2}}=\frac{1}{2\sqrt{\pi}}\int_0^{\infty}\left(1-\prod_{i=1}^n(1+2s a_i)^{-1/2}\right)\frac{ds}{s^{3/2}}.\end{eqnarray*} Since we want to prove that $$J_n\leq \frac{E(\|V\|)}{\sqrt{E(\|V\|^2)}}=E\left(\frac{\|V\|}{(a_1+\cdots+a_n)^{1/2}}\right),$$ from equality (B) without loss of generality we may assume from now on the homogeneity condition $a_1+\cdots+a_n=1.$ With this condition, the inequality $b_1\cdots b_n\leq (\frac{1}{n}(b_1+\cdots+b_n))^n$ for $b_i>0$ applied to $b_i=1+2sa_i$ we get $\prod_{i=1}^n(1+2s a_i)\leq \left(1+\frac{2s}{n}\right)^n$ leading to (A) . For studying the case of equality in (A) , we use the fact that $b_1\cdots b_n= (\frac{1}{n}(b_1+\cdots+b_n))^n$ if and only if $b_1=\ldots=b_n$. Thus we have equality in (A) if and only if $a_1=\ldots=a_n=1/n.$
To prove that $J_n$ is an increasing sequence consider the function
$$f(x)=\frac{1}{2\sqrt{\pi}}\int_0^{\infty}\left(1-\left(1+\frac{s}{x}\right)^{-x}\right)\frac{ds}{s^{3/2}}$$ which satisfies $f(n/2)=J_n.$
Observe that from (C) one has $$\lim_{x\to \infty}f(x)=\frac{1}{2\sqrt{\pi}}\int_0^{\infty}\left(1-e^{-s}\right)\frac{ds}{s^{3/2}}=1.$$
To prove that $f$ is increasing on $(0,\infty)$ enough is to show that for fixed $s$ the function $x\mapsto \left(1+\frac{s}{x}\right)^{-x}$ is decreasing, or that the function $x\mapsto g(x)=x\log \left(1+\frac{s}{x}\right)$ is increasing.This is easy to see since $$g'(x)=\log \left(1+\frac{s}{x}\right)-\frac{s}{s+x}$$ whose sign is the same as the sign of $h(\frac{s}{x})=\left(1+\frac{s}{x}\right)\log \left(1+\frac{s}{x}\right)-\frac{s}{x}.$
Now $h(t)=(1+t)\log(1+t)-t$ is positive for $t>0$ since $h(0)=0$ and $h'(t)=\log(1+t)>0.$
To conclude the proof we get that $$\mathrm{Var}(\|V\|)\leq \left(\frac{1}{J^2_n}-1\right)[E(\|V\|)]^2\leq \left(\frac{1}{J^2_1}-1\right)[E(\|V\|)]^2=\left(\frac{\pi}{2}-1\right)[E(\|V\|)]^2$$ since $J_1\leq J_n$ for all $n$ and since $J_1=E(|Z|)=\sqrt{\frac{2}{\pi}}$ when $Z\sim N(0,1).$ Since $\frac{\pi}{2}-1<1$ we have an improvement of the Kintchine-Kahane inequality, which is in fact optimal.
Of course, all that stuff applies to the Angelo case, which corresponds to $V=\sum g_iX_i$ and $\Sigma=\sum_i X_iX_i^T.$
