# Does $|x|^p$ with $0<p<1$ satisfy the triangular inequality on $\mathbb{R}$?

I am curious about whether $|x|^p$ with $0<p<1$ satisfy $|x+y|^p\leq|x|^p+|y|^p$ for $x,y\in\mathbb{R}$.

So far my trials show that this seems to be right... So can anybody confirm whether this is right or wrong and give a proof or counterexample?

Thank you!!

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The same thing is shown in this answer. See also this question for a generalization of this result. –  Martin Sleziak Apr 21 '12 at 9:46
Thank you @MartinSleziak, that's very informative. –  Vokram Apr 21 '12 at 16:32

By homogeneity, we try to see whether $|1+t|^p\leq 1+|t|^p$ for all $t$, where $0<p<1$. We first look whether it's the case if $t\geq 0$. Define the maps $f(t):=(1+t)^p-t^p-1$ on $\mathbb R_{\geq 0}$. The derivative is $f'(t)=p((1+t)^{p-1}-t^{p-1})\leq 0$, so $f(t)\leq f(0)=0$ and the inequality holds for $t\geq 0$. To deal with the general case, note that $$|1+t|^p\leq |1+|t||^p\leq 1+||t||^p=1+|t|^p.$$

This inequality is useful to deal with the $L^p$ spaces for $0<p<1$, showing that the map $d(f,g)=\int |f-g|^pd\mu$ is a metric.

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Thank you Davide!! And you are right I was exactly trying to prove the Minkovski Inequality for $p<1$ when I bumped into this one:) –  Vokram Apr 21 '12 at 9:15

Since $\frac{d^2 x^p}{dx^2} = p (p-1) x^{p-2} < 0$, the function $x^p$ is a convex function for $x>0$, therefore for $x,y >0$ the inequality is a special case of Jensen's inequality.

$$\frac{x^p+y^p}{2} \geq \left(\frac{x+y}{2}\right)^p =\frac{2}{2^p} \frac{(x+y)^p}{2} \geq \frac{(x+y)^p}{2}$$

For negative $x$ or $y$ one has $|x| + |y| \geq |x + y|$. Therefore

$$\frac{|x|^p+|y|^p}{2} \geq \frac{(|x|+|y|)^p}{2} \geq \frac{|x+y|^p}{2}.$$

Notice that $$1. \quad p-1 < 0 \Rightarrow t^{p-1} >(t+|x|)^{p-1} \quad \forall t \in (0,\infty).$$ Then, integrate 1 from 0 to $|y|$ to obtain the desired inequality.