prove inequality using methods of differential calculus

Question

Could you help me prove following inequality ?
$$(x+y)^{\alpha}\le x^{\alpha} + y^{\alpha} $$ $$x,y\ge 0, \alpha \le 1$$
I don't know from what start, I should use methods of differential calculus.

Answer 1

If $x=0$ or $y=0$, then the inequality is obvious. So let's assume $x,y>0$ and define $u=\frac{x}{x+y}$ and $v=\frac{y}{x+y}$. Then, $u,v\in(0,1)$, and we have $$ \alpha\leq 1, u<1\implies (1-\alpha)\log u\leq 0\implies\log u\leq\log u^\alpha\implies u\leq u^\alpha. $$ Similarly, $v\leq v^\alpha$ and so $1=u+v\leq u^\alpha+v^\alpha$, which is equivalent to the given inequality.

Answer 2

We may assume $0<\alpha<1$, because otherwise the claim is trivial (or wrong). The function $f(x):=x^\alpha$ $(x\geq0)$ is concave, i.e., has a decreasing derivative $f'(x)=\alpha x^{\alpha-1}$. It follows that $$f(x+y)-f(x)=\int_0^y f'(x+t)\>dt\leq \int_0^y f'(t)=f(y)\ ,$$ which is equivalent to the claim.