Prove that $\| v \|_{1} \|v\|_\infty \leq \frac{1+\sqrt{n}}{2} \|v\|^{2}_{2}$

Question

The question asks to prove that

$$\|v\|_{1}\|v\|_\infty \leq \frac{1+\sqrt{n}}{2}\|v\|^2_2.$$

I have shown that $\|v\|_1\|v\|_\infty \leq \sqrt n \|v\|_2 \|v\|_\infty = \sqrt n \|v\|_2^2 \frac{\|v\|_\infty}{\|v\|_2} \leq \sqrt n \|v\|_2^2 $. However, this is not as tight as the desired form. Could someone provide a hint or an answer?

Answer 1

Assume WLOG that all $v_i$'s are non-negative and that $v_1 = \max v_i = \|v\|_\infty$. Your inequality is then equivalent to $$ 2v_1(v_2+\ldots+v_n)\le(\sqrt n - 1)v_1^2 + (\sqrt n+1)(v_2^2+\ldots+v_n^2). $$ Now, for the right-hand side (RHS), \begin{align*} \operatorname{RHS} &= \left(v_1\sqrt{\sqrt n - 1} - \sqrt{\sqrt n + 1}\sqrt{v_2^2+\ldots+v_n^2}\right)^2 + 2v_1\sqrt{n-1}\sqrt{v_2^2+\ldots+v_n^2}\\ &\ge 2v_1\sqrt{n-1}\sqrt{v_2^2+\ldots+v_n^2}\\ &\ge 2v_1(v_2+\ldots+v_n). \end{align*}