# Inequality proof help

I was doing some work and got stuck at a proposition I made to finish it... I just can't seem to find it anywhere... Here it goes: $$a,b>0,r>1\implies \sqrt[r]{a+b}\leq\sqrt[r]{a}+\sqrt[r]{b}.$$

I tried for a plenty of values of $a,b,r$ and it seems to work in all cases... I'd love to see a proof or some counterexamples! Thanks in advance!

Note: $a,b,r$ are real numbers

## migrated from mathoverflow.netMar 2 '14 at 0:00

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• This is not research level. If $r$ is an integer, this is simple algebra. Otherwise, the keyword is "convexity". – Alex Degtyarev Mar 1 '14 at 23:53

It suffices to show that $$1\le \left(\frac{a}{a+b}\right)^{1/r}+\left(\frac{b}{a+b}\right)^{1/r}.$$ But $$0<\frac{a}{a+b},\,\frac{b}{a+b}<1,$$ and as $0<\dfrac{1}{r}<1$, then $$\left(\frac{a}{a+b}\right)^{1/r}>\frac{a}{a+b}\,\,\,\,\,\text{and}\,\,\,\,\, \left(\frac{b}{a+b}\right)^{1/r}>\frac{b}{a+b},$$ and hence $$\left(\frac{a}{a+b}\right)^{1/r}+\left(\frac{b}{a+b}\right)^{1/r}> \frac{a}{a+b}+\frac{b}{a+b}=1.$$
$(\sqrt[r]{a+b})^r \leq (\sqrt[r]{a}+\sqrt[r]{b})^r \implies a+b \leq (\sqrt[r]{a})^r + (\sqrt[r]{b})^r + positiveterms$