# 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?

• In my book it is given like this, by the way the name of the book is "Higher Algebra" by "Hall & Knight". Jun 12 '19 at 12:39

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 $$\mbox{weighted} ~ GM=\left({x_1}^{f_1} {x_2}^{f_2} {x_3}^{f_3}\cdots{x_n}^{f_n}\right)^{\frac{1}{f_1+f_2+f_3+\dots+f_n}}.$$

• Can you please tell me, how do you get that HM? Jun 12 '19 at 13:26
• Yes, $HM=\frac{f_1+f_2+f_3+....+f_n}{\frac{f_1}{x_1}+\frac{f_3}{x_3}+...+\frac{f_n}{x_n}}.$ Jun 12 '19 at 14:11

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 $$a>0.$$

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 $$f(x)=x\ln{x}.$$

• I don't know the Jensen inequality, and it is also not given in my book. So is there any other method than that. Jun 12 '19 at 12:36
• @bigolo Do you know derivatives? Jun 12 '19 at 12:39
• yes, I know differentiation. Jun 12 '19 at 12:40
• @bigolo If so calculate please $(x\ln{x})''.$ What did you get? Jun 12 '19 at 12:41
• the result is $\frac{1}{x}$ Jun 12 '19 at 12:46