Prove that when $a,b \ge 0, r \ge 1$ that $(a + b)^r \ge a^r + b^r$

Question

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!

Answer 1

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!

Answer 2

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?

Answer 3

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 $x\in(0,1)$, $f(x)<1$.