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!

Also, your statement is not quite true. If $a=b=0$ then $(a+b)^{r}=a^{r}+b^{r}=0$ so that inequality is not strict in that case. — pwerth, Jan 25 '19 at 19:26
Moreover, the generalized binomial theorem is not valid for all $a,b \ge 0$ and the mentioned formula is wrong! — p4sch, Jan 25 '19 at 19:28
$r$ is real, your binomial formula does not make much sense. — zwim, Jan 25 '19 at 19:28
@zwim This is generalized binomial formula, valid for any $r$ whenever $|x|>|y|$. — Wojowu, Jan 25 '19 at 19:31
Thank you for fixes, it should have been $\ge$ not $>$ and I did mix up $a,b$ with $x,y$ in the generalized binomial theorem. — clay, Jan 25 '19 at 20:38
Does this answer your question? Can we say that $\left(a+b\right)^{\alpha}>a^{\alpha}+b^{\alpha}$ for all $a,b>0$ and $\alpha>1$? — Martin R, May 06 '20 at 07:07

score 2 · Answer 1 · answered Jan 25 '19 at 19:30

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?

score 2 · Accepted Answer · answered Jan 25 '19 at 19:32

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!

score 1 · Answer 3 · answered Jan 25 '19 at 19:41

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

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

3 Answers3

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