TITLE: How to prove that $2\sqrt{a^{ea}b^{eb}}\ge a^{eb}+b^{ea}$ for $a > 0, b > 0$?
QUESTION [5 upvotes]: Let $a,b\in R^{+}$. Show that
$$2a^{\frac{ea}{2}}b^{\frac{eb}{2}}\ge a^{eb}+b^{ea} \>.$$
My attempt
I know the following inequality is true:
$$a^{ea}+b^{eb}\ge a^{eb}+b^{ea} \>.$$
See this post on AoPS or S. Manyama (2010), Solution of One Conjecture on Inequalities with Power-Exponential Functions, The Australian Journal of Mathematical Analysis and Applications, vol. 7, no. 2.
But my problem is much stronger (because $a^2+b^2\ge 2ab$) and I can't solve it.

REPLY [9 votes]: Let $b=a/2$ :
\begin{align*}\frac{RHS}{LHS} &=  \frac{a^{ea/2}+(a/2)^{ea}}{2 a ^{ea/2 }(a/2)^{e(a/2)/2}} \\ &= \frac{1}{2(a/2)^{ea/4}} + \frac{a^{ea/4}}{2^{1+3ea/4}}\end{align*}
which tends to $\infty$ as $a\to\infty$, because $\displaystyle\frac{a^{ea/4}}{2^{1+3ea/4}} \to \infty$. 
This means that for all large enough $a$, the pairs $(a,b)=(a,a/2)$ are counterexamples.