<s>[INST] Prove that: $x^x\cdot y^y \geq \left(\frac{x+y}{2} \right)^{x+y} $ Prove that:
$$x^x \cdot y^y \geq \left(\frac{x+y}{2} \right)^{x+y} $$
for every $x, y > 0$
Where I got stuck:
$$x^x \cdot y^y \geq \left(\frac{x+y}{2} \right)^{x+y} \iff \ln(x^x \cdot y^y) \geq \ln \left[\left(\frac{x+y}{2} \right)^{x+y} \right] $$
$$x\ln(x) + y\ln (y) \geq (x+y) \ln \left(\frac{x+y}{2} \right) $$
 [/INST] First, rewrite your inequality as $$\frac{x\ln(x)+y\ln(y)}2\geq \left(\frac{x+y}2\right)\ln\left(\frac{x+y}2\right)$$
Now if $f(z)=z\ln z$, then $f'(z)=1+\ln z$ and $f’'(z)=\frac1z$ which is larger than $0$ for $z\in(0,\infty)$. Hence, $z\ln z$ is convex in the domain $(0,\infty)$.
As $x,y\in(0,\infty)$, we can conclude by Jensen's inequality that $$\frac{x\ln(x)+y\ln(y)}2\geq \left(\frac{x+y}2\right)\ln\left(\frac{x+y}2\right)$$
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