Condition for inequality for norms Let $a$ be a vector in $R^m$. Under which condition the following inequality is always true:
$$
\sqrt m \|a\|_{\infty}\leq \|a\|_2^2-\frac{\left(\sum_{i=1}^ma_i\right)^2}{m}.
$$
 A: There is little hope to find a full answer, in particular this is no such answer, but rather a discussion in the hope it might trig some idea to someone else...
Let us choose $x=(1,1,\ldots,1)$ and consider the Cauchy-Schwarz inequality, 
$$\left|\sum x_na_n\right|^2\leq m \sum |a_n|^2$$
This tells us not only that $$0\le \sum|a_n|^2 -\frac{\left|\sum a_n\right|^2}{m}$$
but also that we have equality when $a=\lambda x$, hence in order to have an inequality, like the one in the OP, we need to spot some condition like ``$a$ must be outside of some skew-cone neighbourhood (to be found/defined) of the line $\lambda x$, $\lambda\in\mathbb{R}$''. 
If we look at the case $\mathbb{R}^2$. Consider $a=(x,y)$ where $x<y$, then
$$
\sqrt{2}\|a\|_\infty\leq \|a\|_2^2-\frac{(x+y)^2}{2}
$$
simplifies to 
$$2\sqrt{2}|y|\leq (y-x)^2$$
and then $$x\leq y - 2^{3/4}\sqrt{|y|}.$$
After this one might wish to see some repetition for the $\mathbb{R}^3$ case - but unfortunately that is more a more delicate problem, ending up with 
$$3^{3/2}\max(|x|,|y|,|z|)\leq (x-y)^2 + (x-z)^2+(y-z)^2$$
