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.

  • $\begingroup$ The numbers are positive reals. $\endgroup$ – J. Abraham Nov 17 '18 at 11:52

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.


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 $1$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.