prove inequality using methods of differential calculus 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.  

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.
