Prove or disprove: For any $v\in\mathbb{R}^n, \|v\|_1\|v\|_\infty \leq \frac{1+\sqrt{n}}{2}\|v\|_2^2$

Question

I know that $\|v\|_1\|v\|_\infty \leq (1+\sqrt{n})\|v\|_2^2$, but cannot find a way to refine this bound. Does anyone know where would be a good place to start? Trying out several cases seems to imply that the inequality is true.

Answer 1

Without loss of generality, suppose that the vector $v$ consists of non-negative values $v_1,v_2,\dots,v_n\geq 0$ such that $\|v\|_\infty=v_1$. Therefore it is enough to find a $c$ such that for all these vectors we have: $$ v_1(v_1+\dots+v_n)\leq c(v_1^2+\dots+v_n^2). $$ See that: $$ c(v_1^2+\dots+v_n^2)-v_1(v_1+\dots+v_n)=(c-1)v_1^2+cv_2^2+\dots+cv_n^2-v_1v_2-\dots-v_1v_n. $$ However: $$ \frac{c-1}{n-1}v_1^2+cv_i^2=(\sqrt{\frac{c-1}{n-1}}v_1-\sqrt cv_i)^2+2\sqrt{\frac{c(c-1)}{n-1}}v_1v_i. $$ Then it can be seen that for $c\geq \frac{1+\sqrt n}2$, we have: $$ 2\sqrt{\frac{c(c-1)}{n-1}}v_1v_i-v_1v_i\geq 0. $$ This implies the inequality for all $c\geq \frac{1+\sqrt n}2$.