Prove that Geometric mean is less than Arithmetic mean using Jensen's inequality

Question

So in class we solved the following exercise: state Jensen’s inequality for a convex function and use it to show that for a finite set of real numbers the geometric mean is less than or equal to the arithmetic mean. Unfortunately I lost my notes and cannot recall the proof at all.

Jensen's inequalityis in Royden's book as the following:
let $\phi$ be a convex function on $\mathbb{R}$, $f$ an integrable function over $[0,1]$ and $\phi\circ f$ integrable on $[0,1]$

Then $\phi(\int_0^1 fdx)=\int_0^1 \phi\circ f dx$

This equation seems that to concern the Lebesgue measure however I remember that in class to prove the result we had introduced the counting measure defined on a set. So I guess maybe the above Jensen's inequality is different than the one we used in class.

So does anyone know a proof of the result using the counting measure defined on a set and using some sort of Jensen's inequality possibly different than the one I have above?

Thanks in advance.

Answer 1

Forget about measure theory; I'm pretty sure your class only wants you to use the conceptually simpler result $f(\sum_i w_i x_i)\le \sum_i w_i f(x_i)$, with non-negative $w_i$ summing to $1$. The choice $f(x):=-\ln x,\,w_i=\frac{1}{n}$ gives $-\ln\frac{1}{n}\sum_ix_i\le-\frac{1}{n}\sum_i\ln x_i$. Applying $x\mapsto\exp -x$, which reverses order, completes the proof.

Answer 2

The Jensen's inequality is the result of applying the inequality

$$f(x(1-t)+yt)\le (1-t)f(x)+t f(y)\tag1$$

for some convex function $f$ and $t\in[0,1]$. This can be generalized to

$$f\left(\sum_{k=1}^n a_k x_k\right)\le\sum_{k=1}^n a_k f(x_k)\tag2$$

for $\sum_{k=1}^n a_k=1$ and each $a_k\in[0,1]$. The inequalities above are reverted if $f$ is concave instead of convex.

This can be extended also to series and integrals, in the latter case we can derive the Jensen's inequality.

In your case, for $c_k>0$, we have that

$$\frac1n\sum_{k=1}^n c_k\ge\sqrt[n]{\prod_{k=1}^n c_k}\iff \log\left(\sum_{k=1}^n \frac{c_k}n\right)\ge \sum_{k=1}^n\frac1n\log(c_k)\tag3$$

what holds because $\log$ is concave.