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Prove $(x+y)^a\leq x^a+y^a$ if $0<a\leq1$ and $x,y\geq0$

I need to prove this step for a bigger question. It should be quite basic but I just have no idea...

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    $\begingroup$ If $x+y = 0$, it's clear. Otherwise divide by $(x+y)^a$. $\endgroup$ – Daniel Fischer Feb 5 '14 at 13:31
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Let $f(x)=x^{a}$. We may assume $y\ge x$. The line $y=L(x)$ joining $(0, 0)$ and $(y, f(y))$ satisfies the following $$ f(x)+f(y)\ge L(x)+L(y)=L(x+y)\ge f(x+y), $$ since $f$ is concave.

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The equation is trivial if $x=0$ or $y=0$, so assume neither are $0$.

For $z\ge0$, consider $$ f(z)=(1+z)^a-z^a\tag{1} $$ Note that $f(0)=1$. Furthermore, since $a-1\le0$, $$ \begin{align} f'(z) &=a\left((1+z)^{a-1}-z^{a-1}\right)\\ &\le0\tag{2} \end{align} $$ Thus, for $z\ge0$ and $0\lt a\le1$ $$ (1+z)^a-z^a\le1\tag{3} $$ Set $z=y/x$ and multiply $(3)$ by $x^a$ to get $$ (x+y)^a\le x^a+y^a\tag{4} $$

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