Proving inequality $x^xy^y \geq (\frac{x+y}{2})^{x+y}$ Prove that for all $x,y>0$ the following inequality $x^xy^y \geq (\frac{x+y}{2})^{x+y}$ is true.
It smells like Jensen inequality, but all I can get is that $\frac{x+y}{2}ln(x) + \frac{x+y}{2} ln(y) \geq xln(\frac{x+y}{2})+yln(\frac{x+y}{2})$
 A: This proof won't use Jensen. Multiply both sides by $x^y  y^x$ and regroup:
$$
\begin{alignat*}{}
&\Leftrightarrow &\ x^{x+y} y^{x+y} &\geq{}  \left(\frac{x+y}{2}\right)^{x+y} x^y  y^x\\
&\Leftrightarrow &\ \left(\frac{2xy}{x+y}\right)^{x+y} &\geq{}  x^y  y^x  \\
&\Leftrightarrow &\ \frac{2xy}{x+y} &\geq{}  x^{y/(x+y)} y^{x/(x+y)}
\end{alignat*}
$$
Now apply weighted AM-GM inequality to get:
RHS $\leq \frac{xy}{x+y} + \frac{yx}{x+y} = \frac{2xy}{x+y} =$ LHS. Done.
A: Let $f:(0, \infty) \rightarrow \mathbb{R}$ be given by $x\ln(x)$. $f''(x) = \frac{1}{x} > 0$, therefore $f$ is convex. By Jensen inequality,
$$f\left(\frac{x + y}{2}\right) \leq \frac{1}{2}f(x) + \frac{1}{2}f(y)$$
That is,
$$\left(\frac{x+y}{2}\right)\ln\left(\frac{x + y}{2}\right) \leq \frac{1}{2} x\ln(x) + \frac{1}{2}y\ln(y)$$
That is,
$$\left(\frac{x + y}{2}\right)^{\frac{x + y}{2}} \leq x^{\frac{x}{2}} y^{\frac{y}{2}}$$
Squaring both sides gives the desired result. QED.

Note that in a very similar way, you can show,
$$x^xy^yz^z \geq \left(\frac{x + y + z}{3}\right)^{x+y+z}$$
And even generalize the result further.
