Minkowski's triangle inequality for $0<p<1$

Question

I have a trouble in the following theorem introduced "Minkowski's inequality for $0<p<1$":

Let $0<p<1$ and let $x,y\ge0$. Then, $|x^{p}-y^{p}|\le|x-y|^{p}$.

I have proved the case for $p=\frac{1}{2}$.

How do I prove it in the general case ?

Give some advice! Thank you!

Answer 1

\begin{align*} x^{p}-y^{p}&=\int_{0}^{1}\dfrac{d}{dt}((1-t)y+tx)^{p}dt\\ &=\int_{0}^{1}p((1-t)y+tx)^{p-1}(x-y)dt\\ &\leq(x-y)\int_{0}^{1}p(t(x-y))^{p-1}dt\\ &=(x-y)^{p}, \end{align*} provided that $x\geq y\geq 0$.

Answer 2

A concave function $f:[0,\infty)\to\Bbb R$ such that $f(0)\ge0$ must be subadditive: if by any chance $f(a+b)>f(a)+f(b)$, then $\frac{f(a+b)-f(a)}{b}>\frac{f(b)-f(0)}{b}$, against the hypothesis of concavity.

So, for $p\in(0,1)$ and $x,y\ge 0$, we have that $$(\max\{x,y\}-\min\{x,y\})^p+(\min\{x,y\})^p\ge (\max\{x,y\})^p\\(\max\{x,y\}-\min\{x,y\})^p\ge(\max\{x,y\})^p-(\min\{x,y\})^p\\ \lvert x-y\rvert^p\ge(\max\{x,y\})^p-(\min\{x,y\})^p$$

And, since $x^p$ is an increasing function, the RHS is $\lvert x^p-y^p\rvert$.

Answer 3

If $\,x=y=0\,$ the inequality holds trivially, otherwise assume WLOG that $\,x \ge y\,$ and let $\,t = y/x \in [0,1]\,$. After dividing by $\,x^p \gt 0\,$ the inequality then becomes $\,1-t^p \le (1-t)^p\,$.

Since $\,t^p \ge t\,$ and $\,(1-t)^p \ge 1-t\,$ for $\,0 \lt p \lt 1\,$ and $\,0 \le t, 1-t \le 1\,$, it follows that: $$\,1-t^{p} \;\le\; 1-t \;\le\; (1-t)^{p}\,$$

Answer 4

If for this range of $p$ you already know $x^p+y^p \le (x+y)^p$ for $x,y\ge 0,$ you can do this: Assume WLOG $0<x\le y.$ Then $y^p = (y-x + x)^p \le (y-x)^p + x^p,$ which gives $y^p-x^p \le (y-x)^p.$