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 ?
Thanks in advance for your time . 
Ps: The sign in the identity are the same . 
 A: 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 $~\frac{1}{4}<x<\frac{1}{2}~$ , 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 $
$\hspace{1.7cm} +~68027.7\left(x-\frac{3}{8}\right)^8 + 502855\left(x-\frac{3}{8}\right)^{10} + O\left(\left(x-\frac{3}{8}\right)^{12}\right)$
A: The identity you discovered easily follows from the simple fact that $4=2^2$, which immediately shows that
$$\left(\frac12\right)^{\tfrac18}=\left(\frac14\right)^{\tfrac1{16}}
\qquad\text{ and }\qquad
\left(\frac12\right)^{\tfrac14}=\left(\frac14\right)^{\tfrac18}.$$
