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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?

Thanks in advance.

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  • $\begingroup$ Can we use Minkowski's inequality? $\endgroup$ – Song Feb 16 at 18:33
  • $\begingroup$ yes, if you like! $\endgroup$ – Nikolaos Skout Feb 16 at 18:49
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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.

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