Prove that $f(0)+f(1)\leq f(x)+f(1-x)$

Question

It's a very hard problem that I've worked on for two months:

Let $0\leq x\leq 1$ then define :

$$f(x)=\operatorname{W}\left(x^{\alpha x +1}+1\right)$$

Where $\alpha$ checking: $$2f\left(\frac{1}{2}\right)=\operatorname{W}(1)+\operatorname{W}(2)$$

Then we have :

$$f(0)+f(1)\leq f(x)+f(1-x)\quad(1)$$

To prove it I have tried something like :

$$f(0)\leq f(x)$$

But it really doesn't work to prove $(1)$ !

I have tried also to use Lagrange inversion theorem on the Lambert's function which gives $x\geq 0$:

$$\operatorname{W}(x)=\sum_{n=1}^{\infty}\frac{(-n)^{n-1}}{n!}x^n$$

Without success until now !

Question :

How to show $(1)$ properly ?

Thanks !

Answer 1

With $$f(x)=\operatorname{W}\left(x^{\alpha x +1}+1\right)\qquad \text{and} \qquad 0 \leq x \leq 1$$ because of the symmetry $$f(x)+f(1-x)-f(0)-f(1) \geq 0$$ can only be true if $$2 f\left(\frac{1}{2}\right)-f(0)-f(1) \geq 0$$ that is to say if $$2 W\left(2^{-\frac{\alpha }{2}-1}+1\right)-W(2)-W(1)\geq 0$$ The solution of the equation $$2 W\left(2^{-\frac{\alpha }{2}-1}+1\right)-W(2)-W(1)=0$$is given by $$\alpha_* =-\frac{2 \log \left(e^{\frac{1}{2} (W(1)+W(2))} (W(1)+W(2))-2\right)}{\log (2)}$$ which is $\sim 0.34470677834793192536$.

If $\alpha > \alpha_*$ this is not true for all $x$.

The value of $\alpha_*$ is not recognized by inverse symbolic calculators.

Edit

In order the inequality holds, the first condition is that its first derivative at $x=0$ must be positive. This already gives the condition $$\alpha < \frac{2 W(1)-W(2)+W(1) W(2)}{(W(1)+1) W(2)}\sim 0.572711$$