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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

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migrated from mathoverflow.net Mar 2 '14 at 0:00

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  • $\begingroup$ This is not research level. If $r$ is an integer, this is simple algebra. Otherwise, the keyword is "convexity". $\endgroup$ – Alex Degtyarev Mar 1 '14 at 23:53
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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 $$ \left(\frac{a}{a+b}\right)^{1/r}+\left(\frac{b}{a+b}\right)^{1/r}> \frac{a}{a+b}+\frac{b}{a+b}=1. $$

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  • $\begingroup$ I think I've got it.. I proved that f is concave and that f(0) >= 0 which implies that f is subadditve, meaning that f(x+y) <= f(x) + f(y), am I right? $\endgroup$ – Solero93 Mar 2 '14 at 11:22
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You can raise both terms of the inequality to r. You get:

$(\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 $

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  • $\begingroup$ Thinking about it more, I am not sure how accurate is what I wrote above, for non-integer r. Yiorgos answer is more general. Mine is more straightforward, only if it is correct for non-integer r :) $\endgroup$ – Thanassis Mar 2 '14 at 0:52
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    $\begingroup$ I've thought about that one too, but I'd reckon you're using the binomial theorem there which is only for integer exponents... I'm not quite sure that for non-integer exponents the part $(\sqrt[r]{a})^r + (\sqrt[r]{b})^r would appear... at least I have never heard about this kind of generalization of the binomial theorem... $\endgroup$ – Solero93 Mar 2 '14 at 10:49

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