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!



Inequality for the inner product in the probabilistic simplex
*It seems like there should be a simple proof using standard inequalities*


Most certainly, most certainly. Nate essentially had it though he was doing Jensen on the wrong side.


We have $z\mapsto z^{2c-1}$ convex, so by Jensen
$$
\sum(x\_iy\_i)^c=\sum x\_i(x\_i^{\frac{c-1}{2c-1}}y\_i^{\frac{c}{2c-1}})^{2c-1}\ge\left(\sum x\_ix\_i^{\frac{c-1}{2c-1}}y\_i^{\frac{c}{2c-1}}\right)^{2c-1}=\left(\sum x\_i^{\frac{3c-2}{2c-1}}y\_i^{\frac{c}{2c-1}}\right)^{2c-1}
$$
Similarly, we get
$$
\sum(x\_iy\_i)^c\ge \left(\sum x\_i^{\frac{c}{2c-1}}y\_i^{\frac{3c-2}{2c-1}}\right)^{2c-1}
$$
So,
$$
\sum(x\_iy\_i)^c\ge \sqrt{\left(\sum x\_i^{\frac{3c-2}{2c-1}}y\_i^{\frac{c}{2c-1}}\right)\left(\sum x\_i^{\frac{c}{2c-1}}y\_i^{\frac{3c-2}{2c-1}}\right)}^{2c-1}
\ge \left(\sum x\_iy\_i\right)^{2c-1}
$$
by Cauchy-Schwarz.


