How to show $(a_1a_2\ldots a_n)^{\frac{1}{n}}\leq \frac{\sum_{i=1}^{n}a_i}{n}$

Question

How to show $(a_1a_2\ldots a_n)^{\frac{1}{n}}\leq \frac{\sum_{i=1}^{n}a_i}{n}$ with $a_i$ positive.

Well, I tried by induction: with $n=2$ then $\sqrt{ab}\leq \frac{a+b}{2}$ is equivalent to say (elevate square in both side) $4ab\leq a^2 +2ab + b^2$ and this is equivalent $0\leq(a-b)^2$ and this is true.

I suppose it is true for some $n$. But with $n+1$, I don't know how to do, Please can help me with a hint or other way, thank you so much.

Among the proofs at @Fractal's link, perhaps the simplest proof by induction is here; that one's not famous enough. marty cohen's link contains it too, in much less detail. — J.G., Jun 20 '20 at 17:26

score 1 · Accepted Answer · answered Jun 20 '20 at 17:31

A very simple proof which doesn't require induction (except implicitly); compare the logarithms. You only need to show that $$\frac{\sum_{i=1}^n\ln a_i}n\le\ln\frac{\bigl(\sum_{i=1}^na_i\bigr)}n, $$ and this results from $\ln$ being a concave function.

score 1 · Answer 2 · answered Jun 20 '20 at 17:53

There is a simple proof based on a very nice result here Jensen's inequality, which is very useful in many areas (mathematics, economics, etc).

to understand this result, it suffices to know simple things about convex functions in one variable at the level of High school or first year college Calculus.

How to show $(a_1a_2\ldots a_n)^{\frac{1}{n}}\leq \frac{\sum_{i=1}^{n}a_i}{n}$

2 Answers2