If $a,b,c\ge 0$, $s\in\left(0,1\right)$ and $a^{s}+b^{s}+c^{s} \leq \left( a+b+c\right)^{s}$ then $a,b,c\in${$0,a+b+c$}.
Any hint, please
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2You wrote $a,b,c\in${$0,a+b+c$}. Don't you mean $a,b,c \in (0,a+b+c)$? – Newb Nov 20 '13 at 00:48
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Not really, the problem was stated as such. i.e that a, b and c are either 0 or the sum of all three. – javi Nov 20 '13 at 00:52
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If that's the case, the problem ask to prove that $a^s\leq a^s$ for $a\geq 0$, $s\in (0,1)$, which is ridiculous – leo Nov 20 '13 at 00:54
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1@leo I don't think that's the case – javi Nov 20 '13 at 01:05
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Hint: By Jensen's inequality, if $a, b, c > 0$, then
$$ (a+b+c)^s > a^s + b^s + c^s.$$
Claim in your question follows after filling in some more details.
Calvin Lin
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the question is very strange, $a^x$ is convex so no matter $s$ is less or greater than 1.and $a,b,c$ can be any positive number.there is no limitation for the inequality.what the question really asked for? – chenbai Nov 20 '13 at 06:25
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@chenbai If the inequality holds then a, b and c are either $0$ or the sum of all three. – javi Nov 20 '13 at 13:18
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OK, I made a mistake. indeed, it should use $x^s$, when $0<s<1$, $x^s$ is concave function, so $(a+b+c)^s \le a^s + b^s + c^s$, if you want$(a+b+c)^s \ge a^s + b^s + c^s$, then only case is $(a+b+c)^s = a^s + b^s + c^s \implies a=b=c=0$ – chenbai Nov 21 '13 at 02:40
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