How to apply the AM-GM relation in this inequality?

Here is an inequality that I came across recently.

$((a+b)/2)^m$ $\le$ $(a^m + b^m)/2$, if $m<0$ or $m>1.$

Equality happens when $m = 0$ or $m = 1.$ Inequality reverses if $m$ is between 0 and 1.

I was able to prove this for $m = 2$. Here is what I did:

We know that $(a+b)^2/2 = (a^2 + b^2)/2 + ab.$ ----- (1)

Also, AM of $a^2$ and $b^2$ is greater than or equal to GM of $a^2$ and $b^2.$ That means $(a^2 + b^2)/2$ $\ge$ $ab.$

=> (1) --> $(a+b)^2/2$ $\le$ $(a^2 + b^2)/2 + (a^2 + b^2)/2.$

=> $(a+b)^2/2$ $\le$ $a^2 + b^2.$ Or, $((a+b)/2)^2$ $\le$ $(a^2 + b^2)/2.$

Now, if I try to generalize by replacing '2' with 'm', I'm unable to use the AM-GM inequality to get the desired result.

How to apply the AM-GM inequality in the general case? Or is there any other way to prove the general result?

Solution 1:

For $m>1$ (and $m\in\mathbb{Z}$), assume $0\le a\le b$ and let $c=\frac{a+b}{2}$ and $d=\frac{b-a}{2}$.

Then $a=c-d$ and $b=c+d$.

The LHS is $c^m$, and the RHS is $ \frac{(c-d)^m +(c+d)^m}{2}\ge c^m$ due to any negative terms being cancelled.