Prove $((a+b)/2)^n\leq (a^n+b^n)/2$ Struggling with this proof. 
Prove that $$\left(\frac{a+b}{2}\right)^n≤\frac{a^n+b^n}{2},$$ where $a$ and $b$ are real numbers such that $a+b≥0$ and $n$ is a positive integer.
What technique would you use to prove this (e.g. induction, direct, counter example). How would you go about proving it?
Thanks in advance.

When $n$ is a positive integer, the function $f(x) = x^n$ is convex on $(0, +\infty)$, thus by Jensen's inequality:
$$f\left(\frac{a}{2} + \frac{b}{2}\right) \leq \frac{1}{2}f(a) + \frac{1}{2}f(b),$$
gives the desired inequality.

As pointed out by Andrews, the $f$ defined above is convex on the whole line only when $n$ is even for which case Jensen's inequality can be applied directly. The inequality also holds if both $a$ and $b$ are nonnegative. 
To the case that $n$ is an odd positive integer and $a \leq 0 < b$ (without losing of generality), $a + b \geq 0$, write $n = 2k + 1$, then the right hand side of the inequality is
$$\frac{a^{2k + 1} + b^{2k + 1}}{2} = \frac{1}{2}(a + b)(a^{2k} - a^{2k - 1}b + \cdots + b^{2k}) \geq \frac{1}{2}(a + b)(a^{2k} + b^{2k}) > \frac{a + b}{2}\frac{a^{2k} + b^{2k}}{2}$$
So if we can show 
$$\left(\frac{a + b}{2}\right)^{2k} \leq \frac{a^{2k} + b^{2k}}{2}$$
the inequality holds, but this is the case which we can use Jensen's inequality. 
