# Proving $a^{ab}+b^{ab}\leq a^{a^2}+b^{b^2}$ and an identity .

I was working on this Prove that $a^{ab}+b^{bc}+c^{cd}+d^{da} \geq \pi$ when I have discovered the following identity : $$\Bigg|\Big(\frac{1}{2}\Big)^{\frac{x}{8}}\pm\Big(\frac{1}{4}\Big)^{\frac{x}{8}}\Bigg|=\Bigg|\Big(\frac{1}{2}\Big)^{\frac{x}{4}}\pm\Big(\frac{1}{4}\Big)^{\frac{x}{16}}\Bigg|$$ With $$x$$ a real number such that $$x\neq 0$$

Related to this identity we can have this inequality :

Let $$a,b>0$$ such that $$a+b=\frac{3}{4}$$ and $$\frac{1}{2}\geq a\geq b\geq \frac{1}{4}$$ then we have : $$a^{ab}+b^{ab}\leq a^{a^2}+b^{b^2}$$

I can prove a part of this inequality using log-majorization but I would have a full alternative proof . Furhtermore this inequality is very precise (we have two equality case) and I don't think that reduce to a one variable inequality is a good idea .

So my questions are :

How to prove this two facts ?

• The initial identity is easy to prove, but I really wonder how that is related to the inequality. – Martin R Apr 18 at 19:24
• It would be useful to let the words "can prove a part of this inequality using log-majorization" follow the deeds. Perhaps then we see what you mean. I have the impression that you are on the wrong track. – user90369 Apr 26 at 12:54

The identity is trivial, but the inequality is not trivial. I cannot see any important connection between these two statements.

The following section is just a hint.

Let $$~f(x) := x^{x(\frac{3}{4}-x)} + (\frac{3}{4}-x)^{x(\frac{3}{4}-x)}~$$ and $$~g(x) := x^{x^2} + (\frac{3}{4}-x)^{ (\frac{3}{4}-x)^2}~$$

with $$~\frac{1}{4}\leq x\leq\frac{1}{2}~$$ .

It’s $$~f(\frac{3}{4}-x) = f(x)~$$ , $$~g(\frac{3}{4}-x) = g(x)~$$

and $$~g(x_0) = f(x_0)~$$ for $$~x_0\in\{\frac{1}{4},\frac{3}{8},\frac{1}{2}\}~$$ .

So we are looking for a proof of $$~f(x)\leq g(x)~$$ .

Let $$~\displaystyle G(x) := \frac{g(\frac{1}{2})-g(x)}{(x-\frac{1}{4})(\frac{1}{2}-x)}~$$ and $$~\displaystyle F(x) := \frac{f(\frac{1}{2})-f(x)}{(x-\frac{1}{4})(\frac{1}{2}-x)}~$$ .

Then, equivalent to $$~f(x)\leq g(x)~$$, we should proof:

$$G(x) \leq G\left(\frac{3}{8}\right) = F\left(\frac{3}{8}\right) \leq F(x)$$

This means that we have to find out that $$~G(x)~$$ has it’s (only) maximum

and $$~F(x)~$$ has it’s (only) minimum at $$~x=\frac{3}{8}~$$ :

$$F'(\frac{3}{8})=0~$$ and $$~F''(x)>0$$

$$G'(\frac{3}{8})=0~$$ and $$~G''(x)<0$$

With numerical estimates, we can conclude that $$~F''(x)>4~$$ and $$~G''(x)<-2~$$ for $$~\frac{1}{4} , but it's better to do that by a computer program than by hand.

Note:

Taylor series of $$~G''(x)~$$ at $$~x=\frac{3}{8}~$$ is about:

$$-2.65756 - 25.0727\left(x-\frac{3}{8}\right)^2 - 121.808\left(x-\frac{3}{8}\right)^4 - 792.222\left(x-\frac{3}{8}\right)^6$$

$$\hspace{2cm} -~5121.11\left(x-\frac{3}{8}\right)^8 - 32824.3\left(x-\frac{3}{8}\right)^{10} - O\left(\left(x-\frac{3}{8}\right)^{12}\right)$$

Taylor series of $$~F''(x)~$$ at $$~x=\frac{3}{8}~$$ is about:

$$4.71384 + 122.309\left(x-\frac{3}{8}\right)^2 + 1114.49\left(x-\frac{3}{8}\right)^4 + 8974.12\left(x-\frac{3}{8}\right)^6$$

$$\hspace{1.7cm} +~68027.7\left(x-\frac{3}{8}\right)^8 + 502855\left(x-\frac{3}{8}\right)^{10} + O\left(\left(x-\frac{3}{8}\right)^{12}\right)$$

• @FatsWallers: Thanks for the bounty! ;) – user90369 Apr 29 at 5:00

The identity you discovered easily follows from the simple fact that $$4=2^2$$, which immediately shows that $$\left(\frac12\right)^{\tfrac18}=\left(\frac14\right)^{\tfrac1{16}} \qquad\text{ and }\qquad \left(\frac12\right)^{\tfrac14}=\left(\frac14\right)^{\tfrac18}.$$