TITLE: Elementary proof that $|x|^p$ is convex.
QUESTION [24 upvotes]: I'm writing some notes about analysis and want to use the fact that $|x|^p$ is convex for every $p>1$ to prove Minkowski's inequality. However I didn't wrote anything about derivatives nor limits yet. Is there a simple way to prove this?
EDIT: The only non-trivial inequalities proved yet are the triangular inequality and the Cauchy-Schwarz inequality.

REPLY [2 votes]: Here is an elementary argument which uses only MVT: Claim: for $x \geq 0$ we have $x^{p}=\sup \{a^{p}+(x-a)pa^{p-1}: a\geq 0\}$. To prove that apply MVT to $x^{p}-a^{p}$ and consider the cases $a<x$ and $a \geq x$ separately. [ The idea for this comes from the geometric fact that a convex function is the upper envelope of the tangent lines]. Now define $f_a(x)=a^{p}+(x-a)pa^{p-1}$. [MVT gives $a^{p}+(x-a)pa^{p-1}\leq x^{p}$. To show that the supremum of the left side equals $x^{p}$ you just have to note that $a^{p}+(x-a)pa^{p-1}=x^{p}$ when $a=x$].  Then $f_a(\frac {x+y} {2})=\frac {f_a(x)+f_a(y)} 2$ (since $f_a$ is an affine function ). Just take supremum over $a \geq 0$ to get the desired inequality.