Prove that if $a+b=2$, then $n \in [0..1], \;a^{(2b)^{n}} + b^{(2a)^{n}}\leq 2$

Asked May 25 '17 at 06:30

Active May 25 '17 at 07:24

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Prove that

if $a+b=2$, and $ a,b >0$, $n \in [0..1]$ then $ a^{(2b)^{n}} + b^{(2a)^{n}}\leq 2$

I asked this question because one of a female student in our class (her MathExchange User ID is NH_HN) proved the above inequality while attempting to solve this question. I think the question is interesting and I am curious to know the answer. I posted this question here in the hope that s.o can show me the solution

I have try several value of $n$ between $0$ and $1$. For each value of $n$ I plot a function $f(x) = x^{(2(2-x))^{n}} + (2-x)^{(2x)^{n}}$. It is always less than 2 as the inequality said.

edited May 25 '17 at 07:24

asked May 25 '17 at 06:30

user3375304

1

But before that you need to show your work. – StubbornAtom May 25 '17 at 06:49
@Macavity Sorry it is a big mistake. I have deleted my comment. Thank you very much. – Kelenner May 26 '17 at 08:34

Prove that if $a+b=2$, then $n \in [0..1], \;a^{(2b)^{n}} + b^{(2a)^{n}}\leq 2$

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