Refinement of $a^{4b^2}+b^{4a^2}\leq 1$

Asked Dec 12 '20 at 14:05

Active Dec 12 '20 at 14:05

Viewed 61 times

As I have almost a proof of this fact If $a+b=1$ so $a^{4b^2}+b^{4a^2}\leq1$ .I propose a refinement of this fact because we have :

Let $0<x\leq 0.5$ then we have :

$$x^{4(1-x)^2}+(1-x)^{4x^2}\leq \left(2^{2x+1}x^2(1-x)\right)^{2(1-x)}+\left(1-2^{2x+1}x^2(1-x)\right)^{2x}\leq 1$$

I can show the LHS because we have as first fact :

Let $0<x\leq 0.5$ then we have :

$$x^{2(1-x)}\leq 2^{2x+1}x^2(1-x)$$

And :

$$(1-x)^{2x}\leq 1-2^{2x+1}x^2(1-x)$$

So the part I cannot prove is the RHS :

Have you an idea for that ?

Thanks in advance .

asked Dec 12 '20 at 14:05

Barackouda

1

So you replace an unsolved inequality by a sharper inequality which you cannot solve either? – Martin R Dec 12 '20 at 14:14
@MartinR I'm not agree the inequality of Michael Rozenberg is solved so I propose another one . – Barackouda Dec 12 '20 at 14:18
That's what I said. MR's inequality $a^{4b^2}+b^{4a^2}\leq1$ is unsolved. You ask for a solution to a sharper inequality. – Martin R Dec 12 '20 at 14:22
Should $a$ and $b$ be always positive? Otherwise there may be some troubles, e.g. $a=-1/4$ and $b=5/4$ then $a^{4 b^2} = (-1)^{25/4} = (-1)^{1/4}$ – guest Dec 12 '20 at 15:40
@guest Yes we have $a+b=1$ and $a,b>0$ – Barackouda Dec 12 '20 at 16:12

0 Answers0