If $a+b=1$, show that $a^{4b^2} + b^{4a^2} \leq 1$

Asked Mar 16 '17 at 13:43

Active Mar 16 '17 at 15:11

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If $a+b=1$, show that $a^{4b^2} + b^{4a^2} \leq 1$. It's given that $a,b \in \mathbb{R^+}$.

I couldn't get the equality case actually. I couldn't get a rigorous way to get the equality case.

EDIT: New method For the inequality, WLOG assume $a>b$. Then $a^{4b^2} + b^{4a^2} \leq a^{4a^2} + b^{4a^2} \leq (a+b)^{4a^2} \leq 1$

Is it correct?

Thanks a lot StackExchangers!!

edited Mar 16 '17 at 15:11

asked Mar 16 '17 at 13:43

Mathejunior

$a=b=1/2$, but $a,b \in \mathbb{Z}^+$? – B. Goddard Mar 16 '17 at 13:46
2

Same question (with active bounty): http://math.stackexchange.com/questions/2176571/if-ab-1-so-a4b2b4a2-leq1. – Martin R Mar 16 '17 at 13:47
I have edited it – Mathejunior Mar 16 '17 at 13:47
I do not understand your "expansion method", but I assume that it fails because the exponents are not integers. – Martin R Mar 16 '17 at 13:52
@MartinR O, okay, true that. – Mathejunior Mar 16 '17 at 13:53

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