TITLE: Inequality for the inner product in the probabilistic simplex
QUESTION [8 upvotes]: Let $x, y \in \mathbb{R}^{n}_{\geq 0}$ satisfy $\sum_i x_i = \sum_i y_i = 1$ and let $c \geq 1$. I am trying to find out whether the following inequality holds:
\begin{equation}
\left(\sum_i x_i y_i\right)^{2c-1} \leq \sum_i(x_i y_i)^c
\end{equation}
It seems like there should be a simple proof using standard inequalities, but it eludes me. I need the inequality to hold as part of a larger proof dealing with isoperimetric properties of the boolean hypercube. A proof or a counterexample would be much appreciated!

REPLY [7 votes]: Let me explain how this may be solved without tricks, this may help for other similar issues. 
1) The inequality must hold true for $\sum x_i\leqslant 1$, $\sum y_i\leqslant 1$ (instead of equalities). Indeed, if we replace $x_i$ to $tx_i$, $t>1$, we get a stronger inequality, and this reduces to the case $\sum x_i=1$, analogously to $\sum y_i=1$.
2) Denote $z_i=x_iy_i$. What can be said about these numbers if we know only that $\sum x_i\leqslant 1,\sum y_i\leqslant 1$? For sure, we know that $\sum \sqrt{z_i}=\sum\sqrt{x_iy_i}\leqslant \sum \frac{x_i+y_i}2=1$, but actually nothing more: if $\sum \sqrt{z_i}\leqslant 1$, then it may occur $x_i=y_i=\sqrt{z_i}$. Note that even when we do not see an inequality for $z$'s immediately, we could come to it by minimizing $\sum y_i=\sum z_i/x_i$ subject to $\sum x_i\leqslant 1$.
3) Thus the problem restates as $\sum \sqrt{z_i}\leqslant 1$, we need to prove $(\sum z_i)^{2c-1}\leqslant \sum z_i^c$. Now make the inequality homogeneous, that is, write a (stronger) homogeneous in $z$'s inequality $$\left(\sum z_i\right)^{2c-1}\leqslant \left(\sum z_i^c\right)\left(\sum \sqrt{z_i}\right)^{2c-2}.$$
Ok, after this standard preparation we get a pure Hölder inequality
$$\left(\sum a_i\right)^\alpha\left(\sum b_i\right)^\beta\geqslant 
\left(\sum a_i^{\frac{\alpha}{\alpha+\beta}} b_i^{\frac{\beta}{\alpha+\beta}}\right)^{\alpha+\beta}
$$
for $\alpha=1,\beta=2c-2$, $a_i=z_i^c,b_i=\sqrt{z_i}$.