Prove that when $a,b \ge 0, r \ge 1$ that $(a + b)^r \ge a^r + b^r$ 
Prove that when $a,b,r \in \mathbb{R}, a,b \ge 0, r \ge 1$ that $(a + b)^r \ge a^r + b^r$

My first idea for this proof was to use the generalized binomial theorem:
\begin{align*}
  (a+b)^r &= \sum\limits_{k=0}^\infty \binom{r}{k} a^{r-k} b^k \\
  \binom{r}{k} &= \frac{r \cdot (r - 1) \cdots (r - k + 1)}{k!} \\
\end{align*}
But I'm unable to get that to work. Any suggestions? Thank you!
 A: Hints: You need $a,b>0$ else the inequality is false when one of $\{a,b\}$ is zero. Likewise $r>1$ is needed else the inequality is false when $r=1$. 
So I'll assume $a,b>0$ and $r>1$. Note that for $1>x>0$ and $r>1$, we have:
$$
x^r<x\tag{$*$}
$$ 
(transform this by the strictly increasing function $\log$ and see what happens).
Now, apply ($*$) two times, first with $x=a/(a+b)$ and then with $x=b/(a+b)$. What will happen when you sum up the resulting two inequalities?
A: It should be $$(a+b)^r\geq a^r+b^r.$$
For $ab=0$ it's obvious.
Let $ab\neq0$ and $\frac{a}{b}=x$.
Thus, we need to prove that $f(x)\geq0$, where $$f(x)=(1+x)^r-1-x^r$$
Indeed, $$f'(x)=r\left((1+x)^{r-1}-x^{r-1}\right)\geq0.$$
Thus, $$f(x)\geq f(0)=0$$ and we are done!
A: Let $x=\frac{a}{a+b}$. Then the inequality is equivalent to
$$ x^r+(1-x)^r<1, x\in(0,1)$$
for $r>1$. Let 
$$f(x)=x^r+(1-x)^r<1. $$
So $f'(x)=r[x^{r-1}-(1-x)^{r-1}]=0$, one has $x=\frac12$. Note $f'(x)<0$ if $0<x<\frac12$
and $f'(x)>0$ if $\frac12<x<1$. Thus $f(x)$ is decreasing in$[0,\frac12]$ and $f(x)$ is increasing in$[\frac12,0]$ and hence $f(x)$ reaches the maximal value $1$ either at $x=0$ or at $x=1$. Thus for $x\in(0,1)$, $f(x)<1$.
