If $a_1,\dots,a_n$ are all unequal positive quantities, then prove that:
$$\prod_{i=1}^n a_i^{a_i} > \left(\frac{\sum_{i=1}^n a_i}{n}\right)^{\sum_{i=1}^n a_i}$$
No other conditions are given.
I tried to solve it using logarithms, but I could not understand how can I prove $${a_1\log a_1}+{a_2 \log a_2}+\dots+{a_n\log a_n} > {(a_1+\dots +a_n)}
\left(\log(a+\dots+a_n)-\log(n)\right)$$
And please tell me, is there any standard method to solve this kind of problems?
Best Answer
Another way.
Easy to see that our inequality does not depend on the substitution $a_1\rightarrow ta_1,\dots,a_n\rightarrow ta_n$, where $t>0$.
Thus, we can assume that $a_1+a_2+\dots+a_n=n$ and we need to prove that $$\sum_{cyc}a\ln{a}\geq0$$ or $$\sum_{cyc}(a\ln{a}-a+1)\geq0,$$ which is true because $$a\ln{a}-a+1\geq0$$ for all $a>0.$