How do I show that $a^p + b^p > (a + b)^p$ for $a,b>0$ and $0<p<1$? [closed]

A question on my math homework asks us to show that if $0 < p < 1$ and $a, b > 0$, then $a^p + b^p > (a + b)^p$. I have no idea how to do this, any pointers?

First, the inequality is homogeneous, so you can divide both $a$ and $b$ by $a+b$, then you have to show that $$x^p+y^p>1$$ if $x+y=1$, $x,y>0$ and $0<p<1$. But now, $x,y<1$ so $x^p>x$ and $y^p>y$ for $0<p<1$. And you are done.

Fix $a\gt 0$, and consider $f(x)=a^p+x^p-(a+x)^p$. Note that $f(0)=0$. We have $f'(x)=x^{p-1}-(a+x)^{p-1}$. Since $p-1\lt 0$, we have $f'(x)\gt 0$ whenever $x\gt 0$. So $f(x)$ is increasing, and the result follows.

Setting $$t=\frac a{a+b},$$ you can rewrite $$t^p+(1-t)^p>1,$$ with $0<t,1-t<1$.

For $0<p<1$,

$$t^p>t\text{, and }(1-t)^p>1-t.$$ By summing you get the desired result.