Prove that $|x^a-y^a| \le |x-y|^a$ $if$ $0\le a \le 1 $, $0\le x$ $and$ $0\le y$.

Question

I found this inequality in a test, and while it was easy to prove it if a = 1/2, I didn't manage to prove it for $0\le a \le 1 $. I know that this proof as nothing to do with the triangle inequality, in fact the problem is nearly identical if we suppose that $0\le y \le x $. I'd appreciate any help.

Answer 1

Without loss of generality, we may assume that $x\ge y$. Dividing both sides by $y^a$ and taking all terms to the right, we want to show that $$f(z)=(z-1)^a-z^a+1\ge0\quad\forall z>1$$ $\left(\text{where }z=\dfrac xy\right)$Differentiating with respect to $z$, we get, $$f'(z)=a(z-1)^{a-1}-az^{a-1}\ge0\quad\forall z>1$$ (since $a-1\le0$). Also, $$f(1)=0$$ Therefore, $f$ is a monotonic increasing function with $1$ as a root. Hence, it takes positive values $\forall z>1$ and we are done.