Prove the inequality $x^\alpha \le y^\alpha + z^\alpha$. Given a triplet of non-negative numbers $x$, $y$, $z$ for which holds $x \le y + z$
one needs to prove that the inequality $x^\alpha \le y^\alpha + z^\alpha$ is also correct for all $\alpha \in (0;1)$.
I feel I'm missing something obvious, but can't quite figure it out. I hoped to make use of Minkowski inequality...
 A: If $x = 0$, then it's obvious. Otherwise, dividing both sides of the first inequality by $x$ and second by $x^\alpha$, it's equivalent to showing that if $y + z \geq 1$, then $y^\alpha + z^\alpha \geq 1$. 
We have two cases:


*

*If one of the $y, z$ is greater or equal to $1$, say $y \geq 1$, then also $y^\alpha \geq 1$, so $x^\alpha + y^\alpha \geq 1$

*Otherwise, if both $x, y < 1$, then $x^\alpha \geq x, y^\alpha \geq y$, so $x^\alpha + y^\alpha \geq x + y \geq 1$.

A: It is sufficient to show $(x+y)^{\alpha} \leq x^{\alpha} + y^{\alpha}$ for any $x,y\geq 0$. Suppose $0 <y\leq x$. Use the fact that the function $\frac{x^{\alpha}}{x}$ is decreasing to deduce $\frac{(x+y)^{\alpha}}{x+y} \leq \frac{x^{\alpha}}{x}$. Multiply both sides of the inequality by $(x+y)$ to get $(x+y)^{\alpha} \leq x^{\alpha}(1+\frac{y}{x})$. Since $\frac{y}{x} \leq 1$, we have $\frac{y}{x} \leq (\frac{y}{x})^{\alpha}$ and we are done. The other case with $0 < x \leq y$ can be handled similarly. The trivial cases with either $x=0$ or $y=0$ can easily be handled.
