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Find least upper bound of $\{a^{2018} + b^{2018} + c^{2018} | a + b + c = 1, a, b, c > 0 \}$. I tried power mean inequality, but could only find greatest lower bound.

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  • $\begingroup$ The numbers are positive reals. $\endgroup$ – J. Abraham Nov 17 '18 at 11:52
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Let $f(x)=x^{2018},$ where $x\geq0$ and $a\geq b\geq c\geq0.$

Thus, since $(a+b+c,0,0)\succ(a,b,c)$ and $f$ is a convex function, by Karamata we obtain: $$1=(a+b+c)^{2018}+0^{2018}+0^{2018}\geq a^{2018}+b^{2018}+c^{2018}.$$ The equality occurs for $a=1$ and $b=c=0$.

In our case our variables are positives, which says that $1$ is a supremum.

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Well, $0<a,b,c<1$, so $a^{2018}<a$ etc., therefore $a^{2018}+b^{2018}+c^{2018}<a+b+c=1$. I reckon one can get close to $1$.

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  • $\begingroup$ and a lower bound ? $\endgroup$ – Henno Brandsma Nov 17 '18 at 12:00
  • $\begingroup$ Can you prove that it is a least upper bound? $\endgroup$ – J. Abraham Nov 17 '18 at 12:05
  • $\begingroup$ @J.Abraham All numbers in the set are $\le 1$, and $1$ is assumed for $(a,b,c)=(1,0,0)$, e.g. $\endgroup$ – Henno Brandsma Nov 17 '18 at 12:17
  • $\begingroup$ @Henno $(1,0,0)$ is not a valid solution since we require all numbers to be greater than $0$. $\endgroup$ – Sambo Nov 17 '18 at 12:38
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    $\begingroup$ @Sambo true, use $t,t,1-2t$ for small $t$ instead. We can get as close as we like to $1$. $\endgroup$ – Henno Brandsma Nov 17 '18 at 12:40

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