How to prove the inequality $a_1^{a_1}a_2^{a_2}\cdots a_n^{a_n} > \left(\frac{a_1+\dots+a_n}{n}\right)^{a_1+\dots+a_n}$? 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?
 A: 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 $a>0.$
A: The hint:
Write it so:
$$\frac{a_1\ln{a_1}+\dots+a_n\ln{a_n}}{n}\geq\frac{a_1+\dots+a_n}{n}\ln\frac{a_1+\dots+a_n}{n}$$
and use  Jensen for $f(x)=x\ln{x}.$
A: Let  $n$ positive elements $a_1,\dots,a_n$ be the variables and $a_1,\dots,a_n$ are their respective frequencies. Then GM of this data is
$$GM=\left(a_1^{a_1}a_2^{a_2}\cdots a_n^{a_n}\right)^{\left(\frac{1}{a_1+\dots+a_n}\right)}$$ and
the corresponding $HM$ is
$$HM=\frac{a_1+a_1+a_3+\dots+a_n}{\frac{a_1}{a_1}+\frac{a_2}{a_2}+\frac{a_3}{a_3}+\dots+\frac{a_n}{a_n}}=\frac{a_1+a_2+a_3+\dots+a_n}{n}.$$
The required result follows from the fact that wieghted GM $\ge$ weighted HM.
$$\mbox{weighted HM}=\frac{f_1+f_2+f_3+\dots+f_n}{\frac{f_1}{x_1}+\frac{f_2}{x_2}+\frac{f_3}{x_3}+\dots+\frac{f_n}{x_n}}$$ and $$\mbox{weighted} ~ GM=\left({x_1}^{f_1} {x_2}^{f_2} {x_3}^{f_3}\cdots{x_n}^{f_n}\right)^{\frac{1}{f_1+f_2+f_3+\dots+f_n}}.$$
