<s>[INST] Nice power sum inequality $ab(a-b)\leq a^{ab}-b^{ab}$ for $a+b=2$ 
Let $a\geq b>0$ such that $a+b=2$ then we have :
  $$ab(a-b)\leq a^{ab}-b^{ab}$$

My try :
We study the function $f(x)$ on $[0;1]$ such that :
$$f(x)=(2-x)^{((2-x)x)}-x^{((2-x)(x))}-(2-x)x(2-2x)$$
the derivative is equal to :
$$f'(x)= x^{(x (2 - x))} (-((2 - x) + (2 - 2 x) \log(x))) + (2 - 2 x) x + 2 (2 - x) x - (2 - 2 x) (2 - x) + (2 - x)^{(x (2 - x))} ((2 - 2 x) \log(2 - x) - x)$$
But I can't show that $f'(x)\geq0$
Any hints would be appreciable. 
Thanks in advance.
Edit :
If we study the function $f(x)$:
$$f(x)=\frac{a^x-b^x}{a-b}$$
The function is increasing and the minimum is reached for $a=b=1$
So we study the following limit :
$$\lim_{a,b \to 1}\frac{a^x-b^x}{a-b}$$
This limit is equal :
$$\lim_{a,b \to 1}\frac{a^x-b^x}{a-b}=x$$
Put : $x=ab$ we have the result . 
 [/INST] Let $a=1+x$ and $b=1-x$.
Thus, $x\geq0$ and
$$a^{ab}=(1+x)^{1-x^2}=1+x-x^3-\frac{x^4}{2}+\frac{1}{6}x^5+\frac{5}{12}x^6+O(x^7)$$ and
$$b^{ab}=(1-x)^{1-x^2}=1-x+x^3-\frac{x^4}{2}-\frac{1}{6}x^5+\frac{5}{12}x^6+O(x^7).$$
Can you end it now? 
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