Hello!


I want to prove that $x = 0.5$ is the global maximum of the function


$f(x) = \frac{(1-a)^2e^{(2x\cdot(x-1)a^2)}}{(1-a)(e^{(2x\cdot(2x-1)a^2)}+e^{((2x-1)\cdot(2x-2)a^2)})-2(1-2a)e^{(4x\cdot(x-1)a^2)}}$ 


where $a\in(0,1)$ and $x\in[0,1]$. I tried to show this using "conventional" ways, but unfortunately the attempts have failed.


Thanks a lot!



Global maximization of a particular function
I have to run out now, so will write only the hint as of now. Later I might complete the details. Let me do one case here, the other one probably follows similarly. 


Let $a \in (1/2,1)$, and write $a=1/2 + \epsilon$, for an appropriate $\epsilon$.


Now consider the function $1/f(x)$, which is (after some cleanup) seen to be


\begin{equation\*}
\frac{2 e^{-2 (-1+x) x \left(\frac{1}{2}+\epsilon \right)^2} \left(e^{2 (-1+x) (-1+2 x) \left(\frac{1}{2}+\epsilon \right)^2} (1-2 \epsilon )+e^{2 x (-1+2 x) \left(\frac{1}{2}+\epsilon \right)^2} (1-2 \epsilon )+8 e^{(-1+x) x (1+2 \epsilon )^2} \epsilon \right)}{(1-2 \epsilon )^2}
\end{equation\*}


But this is a nonnegative (since $2\epsilon < 1$) combination of convex functions of $x$ (because we have exponentials, products are ok), and is thus itself convex. Now, you can numerically or directly optimize this function and see that its minimum occurs at $x=0.5$, which implies that the maximum of your function occurs at $x=0.5$. (assuming it is easy to guarantee that $f(x)$ is a positive function).


