不等式小記

Problem (Austrian Mathematical Olympiad 2008). Prove that the inequality $ \displaystyle \sqrt{a^{1-a}b^{1-b}c^{1-c}}\leq \frac{1}{3}$ holds for all positive real numbers $  a,b,c$ with $  a+b+c=1$.

Those who always discuss math with me will know how to solve it, that's not hard. Now it can be easily generalized to \[ \sqrt{a_1^{1-a_1}a_2^{1-a_2}\cdots a_n^{1-a_n}}\leq \frac{1}{n^{(n-1)/2}}\] with $  a_1+a_2+\cdots +a_n=1$. From this we get for any $  a_i>0$,\[\frac{a_1+a_2+\cdots+a_n}{n}\ge \left(\frac{a_1a_2\cdots a_n}{(a_1^{a_1}a_2^{a_2}\cdots a_n^{a_n})^{1/\sum_{i=1}^na_i}}\right)^{1/(n-1)}.\]
We have established an extremely ugly lower bound. This inequality is useful if we are given a condition on $  \prod a_i^{a_i}$, for example from now on I can show that if \[\boxed{x^xy^yz^z=1, x,y,z>0},\] then \[ \sum_{cyc}\sqrt{\frac{x}{yz}}\ge 3.\]
Take $n =2$, set $  (a_1,a_2)=(x,y)$. The inequality reduces to \[\tfrac{1}{2}(x+y)\ge (x^yy^x)^{1/(x+y)}\iff \left(\tfrac{1}{2}(x+y)\right)^{x+y}\ge x^yy^x
\] (at least I think this inequality is somehow useful), by a little bit argument the following holds for all positive $ x,y$, \[
\sqrt[m]{\frac{x^m+y^m}{2}}\mathop{\ge}\limits_{m\ge n}\sqrt[n]{\frac{x^n+y^n}{2}}\mathop{\ge}\limits_{n\in\mathbb{N}}\frac{x+y}{2}\ge \sqrt{xy}\ge \frac{2}{\frac{1}{x}+\frac{1}{y}}\ge (x^yy^x)^{1/(x+y)}.\]