I originally wanted to compute the following quantity $$ \langle v_{\max.}^2 \rangle_{S_d} = \left\langle \frac{\| x\|_{\infty}^2}{\|x\|_2^2} \right\rangle_{\mathcal N_n(0,1)} $$ which is the average value of the squared ratio between the max norm and the 2-norm, where $$ x_i \sim \mathcal N(0,1)$$ This is basically the average value of the max of a vector over the $n$-sphere. I managed to relate this quantity to the fact that this ratio converges to a constant or to zero: $$ \frac{E[\|x\|_2^2\cdot\|x\|_\infty^{2}]}{E[\|x\|_2^2]E[\|x\|_\infty^{2}]}=\frac{E[\|x\|_2^2\cdot\|x\|_\infty^{2}]}{2n\ln(n)} $$
I know that the upper bound could be $$ \frac{E[\|x\|_2^4]}{2n\ln(n)}=\frac{n(n+2)}{2n\ln(n)}=\frac{(n+2)}{2\ln(n)} $$ but I also know that this is a rough approximation from numerical simulations since it diverges and doesn’t tend to a constant. Could someone help me find better upper bounds for this or just directly solve the initial average of the ratio?