Convexity of $x^p$ -- without calculus At least in courses in Germany, you define $\exp(x)=\sum_{n=0}^\infty x^n/n!$ much earlier than establishing basic calculus. Nevertheless, one can easily prove the convexity of $\exp$: For $t\in[0,1]$ and $x>y$ just estimate
$$
\exp(t(x-y))-1 =\sum_{n=1}^\infty t^n(x-y)^n/n! 
\le t\sum_{n=1}^\infty (x-y)^n/n! = t(\exp(x-y)-1),
$$
and it remains to multiply this inequality by $e^y$.
Is there an equally "elementary" proof for the convexity of $x^p=\exp(p\log(x))$
for $p>1$?
 A: It can be shown that if $f(x), g(x)$ are positive, convex and increasing, then $f(x)g(x)$ also is. Noting that $f(x)=x$ is convex, we can inductively find $f(x)=x^p$ to be convex for all $p\in\mathbb{N}$.
Proof: Suppose $f,g$ are convex, $x,y,\in\mathbb{R}$, $\lambda\in[0,1]$. Then
$$f((1-\lambda)x+\lambda(y))g((1-\lambda)x+\lambda(y))\leq[(1-\lambda)f(x)+\lambda f(y)][(1-\lambda)g(x)+\lambda g(y)]$$
Upon expanding, we see this is:
$$(1-\lambda)^2f(x)g(x)+\lambda(1-\lambda)f(x)g(y)+\lambda(1-\lambda)f(y)g(x)+\lambda^2f(y)g(y)$$
$$=(1-\lambda)^2f(x)g(x)+\lambda(1-\lambda)f(x)g(x)+\lambda(1-\lambda)f(x)[g(y)-g(x)]+\lambda(1-\lambda)f(y)g(y)+\lambda(1-\lambda)f(y)[g(x)-g(y)]+\lambda^2f(y)g(y)$$
$$=(1-\lambda)f(x)g(x)-\lambda(1-\lambda)[f(x)-f(y)][g(x)-g(y)]+\lambda f(y)g(y)$$
The condition that they both be increasing gives that the middle term is negative, whence the whole expression is $\leq(1-\lambda)f(x)g(x)+\lambda f(y)g(y)$, as required.
Still working on something for the other $p$.
