About QM power inequality.

Question

It is easy to see that $$x^xy^y \geq\left(\frac{x+y}{2}\right)^{x+y}$$ for all positive real numbers.

but How to prove this inequality? $$x^xy^y \geq\left(\sqrt\frac{x^2+y^2}{2}\right)^{x+y}$$ help me please.

Answer 1

Since our inequality is symmetric, we can assume that $x\geq y$ or $x=ty$, where $t\geq1$.

Thus, we need to prove that $$(ty)^{ty}y^y\geq\left(\frac{t^2y^2+y^2}{2}\right)^{\frac{ty+y}{2}}$$ or$$t^t\geq\left(\frac{t^2+1}{2}\right)^{\frac{t+1}{2}}$$or $f(t)\geq0$, where $$f(t)=\frac{2t}{t+1}\ln{t}-\ln(t^2+1)+\ln2.$$ Now, $$f'(t)=\frac{2\ln{t}}{(t+1)^2}+\frac{2}{t+1}-\frac{2t}{1+t^2}=\frac{2\ln{t}}{(t+1)^2}-\frac{2(t-1)}{(t+1)(1+t^2)}=\frac{2g(t)}{(t+1)^2},$$ where $$g(x)=\ln{t}-\frac{t^2-1}{t^2+1}.$$ We see that $$g'(t)=\frac{1}{t}-\frac{4t}{(t^2+1)^2}=\frac{(t^2-1)^2}{t(t^2+1)^2}\geq0,$$ which says that $g(t)\geq g(1)=0$. Id est, $$f(t)\geq f(1)=0$$ and we are done!