Nice power sum inequality $ab(a-b)\leq a^{ab}-b^{ab}$ for $a+b=2$

Question

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 .

Answer 1

We may assume $a> b$ without loss of generality and write $$ \frac{a^{ab}-b^{ab}}{a-b}=\frac{ab}{a-b}\int_b^a t^{ab-1}\mathrm dt. $$ Since $\sqrt{ab}< \frac{a+b}2=1$, $f(t)= t^{ab-1}$ is convex, hence for all $u\ge 0$ $$ f(u)\ge f(1) +f'(1)(u-1)=1+f'(1)(u-1). $$ Integrating over $[b,a]$ yields $$ \int_b^a t^{ab-1}\mathrm dt\ge \int_b^a 1+f'(1)(t-1)\ \mathrm dt=a-b $$ hence giving $$ \frac{a^{ab}-b^{ab}}{a-b}\ge ab $$ as wanted.

Answer 2

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?