Prove that $f(x)$ have a maximum at $x=0.5$ for $0\leq x\leq 1$

Question

It's an idea that I cannot substantiate. I want to show :

Let $a,b>0$ such that $a+b=1$ and $0\leq x\leq 1$ then we have : $$f(x)=(ax+b(1-x))^{2(a(1-x)+bx)}+(bx+a(1-x))^{2(b(1-x)+ax)}=g(x)+h(x) \leq 1.$$

So we examine :

$$f(x)=(ax+b(1-x))^{2(a(1-x)+bx)}+(bx+a(1-x))^{2(b(1-x)+ax)}=g(x)+h(x)$$

The problem is :

$$f(x)\leq f(\frac{1}{2})$$

So the idea is : prove that $x=0.5$ is a maxima .

Following a the natural way we differentiate $f(x)$ and My claim is the following : the derivative of $f(x)$ is positive for $0\leq x \leq 0.5$ and negative for $0.5\leq x \leq 1$ .

Inspecting the behavior of $f(x)$ it seems that for some value $a,b$ (for example $a=0.6$) the functions $g'(x)$ or $h'(x)$ have an maxima on $I=[0,1]$ . I mean in fact that on a part of this interval $I$, $g'(x)$ is increasing while $-h'(x)$ is decreasing wich is an advantage . Remains to examine the second derivative of $g'(x)$ or $h'(x)$ to find this local extrema . Maybe we can use Lambert's function .

Also to prove the claim we can use a line such that :$f'(x)\leq ux+v$ and $f'(x)\geq ux+v$ if we are on $[0.5,1]$ or $[0,0.5]$.

Update 30/10/2020 :

For $0\leq x\leq 1$ and $0.6\geq a \geq b \geq 0.4$ such that $a+b=1$ we have the following approximation :

$$g(x)=(ax+b(1-x))^{2(a(1-x)+bx)}\simeq p(x)=x(g(1)-g(0))+g(0)$$

$$h(x)=(bx+a(1-x))^{2(b(1-x)+ax)}\simeq q(x)=x(h(1)-h(0))+h(0)$$

So we have :

$$f'(x)=g'(x)+h'(x)=g(x)u(x)+h(x)v(x)\simeq p(x)u(x)+q(x)v(x)$$

Update 31/10/2020

I start with a particulare case let $a=0.6$ and $b=0.4$

First fact $(1)$ :$0.33<0.4^{1.2}$

Second fact $(2)$ :

We have for $0\leq x \leq 0.5$

$$l(x)=2x(0.5-0.33)+0.33\leq (0.6 x+(1-x)0.4)^{2 (0.4 x+(1-x)0.6)}$$

Third fact $(3)$ :

$$0.67>0.6^{0.8}$$

Fourth fact $(4)$ :

We have for $0\leq x \leq 0.5$ :

$$t(x)=2x(0.5-0.67)+0.67\geq (0.4 x+(1-x)0.6)^{2 (0.6 x+(1-x)0.4)}$$

With facts $(2)$ and $(4)$ we have fact $(5)$ for $0\leq x \leq 0.5$:

$$f'(x)=g'(x)+h'(x)=g(x)u(x)+h(x)v(x)\geq g(x)l(x)+t(x)v(x)\geq 0$$

My question: How to prove the claim?

Answer 1

And why not substituting directly $b = 1-a$? If we do that, we want to find the global maximum for $f(a) = a^{2(1-a)} + (1-a)^{2a} = e^{2(1-a) \ln a}+e^{2a \ln (1-a)} $. If you plot this function, you'll see that the maximum is attained at $a = 0, \frac 12, 1$. ($f(0)$ and $f(1)$ are to be understood as a continuous extension).