# Without use of derivatives, prove that function $(1+x^p)^{1/p}$ is convex for $p\geq 1$

Without use of derivatives, prove that function $$(1+x^p)^{1/p}$$ is convex for $$p\geq 1.$$

Attempt. The result is obvious for $$p=1$$, since function $$x+1$$ is affine. For $$p>1$$, functions $$1,~x^p$$ are both convex and $$x\mapsto \sqrt[p]{x}$$ is increasing but not convex (if fact it is concave), in order to use the composition theorem:

$$(convex ~\&~ increasing)\circ convex=convex.$$

Regarding the definition, due to the function being continuous, it is enough to prove mid-point convexity. We have to prove that for all $$x,~y:$$

$$\left(1+\left(\frac{x+y}{2}\right)^p\right)^{1/p}\leqslant\frac{(1+x^p)^{1/p}+(1+y^p)^{1/p}}{2}$$

which after some basic calculations leads to:

$$2^p+2^{p-1}(x+y)^p\leqslant \big((1+x^p)^{1/p}+(1+y^p)^{1/p}\big)^p.$$

How could one procceed?

Let us show mid-point convexity, which is equivalent to convexity for continuous $$f$$. (In fact, the method can establish convexity as well.) Let $$x,y\ge 0$$ be given and let $$\mathbf a, \mathbf b\in \Bbb R^2$$ be $$\mathbf a = (1,x)\ \ \ \text{ and } \ \ \ \mathbf b=(1,y).$$ If we denote $$\|(x_1,x_2)\|_p =(x_1^p+x_2^p)^{\frac1{p}}$$ by $$p$$-norm on $$\mathbb R^2$$, then Minkowski's inequality says that it holds $$\left\|\frac{\mathbf a+ \mathbf b}{2}\right\|_p\ \le \frac12\|\mathbf a\|_p +\frac12 \|\mathbf b\|_p.$$ Since $$\frac12(\mathbf a+\mathbf b) = (1,\frac{x+y}2)$$, it follows that $$\left(1+\left(\frac{x+y}2\right)^p\right)^{1/p}\le \frac12\left(1+x^p\right)^{1/p}+ \frac12\left(1+y^p\right)^{1/p},$$ that is, $$f\left(\frac{x+y}2\right)\le \frac{f(x)+f(y)}2.$$ This implies $$f$$ is a convex function.