Find least upper bound of $\{a^{2018} + b^{2018} + c^{2018} | a + b + c = 1, a, b, c > 0\}$

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.

• The numbers are positive reals. – 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, $$0, so $$a^{2018} etc., therefore $$a^{2018}+b^{2018}+c^{2018}. I reckon one can get close to $$1$$.

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