I have this inequality but I am unsure how to prove it:
0$<\alpha \leq$1
a$^\alpha$+b$^\alpha$ $\geq$ (a+b)$^\alpha$
$\forall a,b \geq 0$
I was given a hint: we can assume b$\geq$0 $(\frac{a}{b})^\alpha$+1 $\geq$ $(\frac{a}{c}+1)^\alpha$ so it suffices to prove f(x)$\geq$ 0:
f(x)=x$^{\alpha}$+1 - (x+1)$^\alpha$
We have equality if $\alpha=1$. So we can assume that $\alpha\lt 1$. We have equality when $a=b=0$. So without loss of generality we can assume that $b\gt 0$. Then our inequality is equivalent to $$\left(\frac{a}{b}\right)^\alpha+1\ge \left(\frac{a}{b}+1\right)^\alpha.$$ We will show that $$x^\alpha +1\ge (x+1)^\alpha\quad\text{when}\quad x\ge 0.\tag{1}$$ Setting $x=\frac{a}{b}$ will yield the desired result.
Let $f(x)=x^\alpha +1-(1+x)^\alpha$. To prove Inequality (1), we will show that $f(x)\ge 0$ for all $x\ge 0$.
We have $f(x)=0$ when $x=0$. We will show that for $x\gt 0$, $f(x)$ is an increasing function. That implies that $f(x)\gt 0$ for all $x\gt 0$. Note that $$f'(x)=\alpha\left(\frac{1}{x^{1-\alpha}}-\frac{1}{(1+x)^{1-\alpha}}\right).$$ Thus $f'(x)$ is positive if $x\gt 0$, since $x^{1-\alpha}\lt (1+x)^{1-\alpha}$. It follows that for $x\gt 0$, $f(x)$ is an increasing function. Since $f(0)=0$, this completes the proof.
