Convex function on the graph but impossible to prove??!

Question

How can we show that the function $f:\left (0,a\right )\to (0,1)$ given by

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

is convex on $\left (0,\frac{a}{2}\right )$?

I made the graph and I computed its first derivative:

$$f'(x)=\dfrac{a\left (\dfrac{1}{x^2}+\dfrac{1}{(a-x)^2}\right )e^{\frac{a}{x}-\frac{a}{a-x}}}{\left (1+e^{\frac{a}{x}-\frac{a}{a-x}} \right )^2}$$.

Now we have to show that $f'$ is increasing on $\left (0,\frac{a}{2}\right )$. I computed the second derivative and even the third derivative (they are too ugly to post them here: I used https://www.derivative-calculator.net). What I show is that $x=a/2$ is a solution of $f''(x)=0$. The graph of $f'$ shows that $a/2$ is the unique maximum point.

I am stucked in proving that the equation:

$$\left (\dfrac{2}{(a-x)^3}-\dfrac{2}{x^3}\right )\cdot \left (e^{\frac{a}{a-x}-\frac{a}{x}}+e^{\frac{a}{x}-\frac{a}{a-x}}+2 \right )=a\left (\dfrac{1}{x^2}+\dfrac{1}{(a-x)^2} \right )^2\cdot \left (e^{\frac{a}{x}-\frac{a}{a-x}}-e^{\frac{a}{a-x}-\frac{a}{x}} \right )$$ has a unique solution $x\in (0,a)$ and that solution is $x=\dfrac{a}{2}$. It is obviously true from the graphic but seems impossible to prove. I feel very frustrated because of that. Maybe there is some trick...or other way of doing it.

Answer 1

Let $g(x)$ be a $C^2$-function with $g'(x) \neq 0$ everywhere, and let $f(x) = \frac{1}{1+\exp(g(x))}$. Then

\begin{align*} f''(x) \geq 0 &\iff \frac{f''(x)}{f(x)^3} \geq 0 \\[0.5em] &\iff e^{g(x)} ((e^{g(x)}-1) g'(x)^2 - (e^{g(x)}+1) g''(x)) \geq 0 \\[0.5em] &\iff \frac{e^{g(x)}-1}{e^{g(x)}+1} \geq \frac{g''(x)}{g'(x)^2}. \tag{1} \end{align*}

Now we let $g(x) = \frac{1}{x}-\frac{1}{1-x}$, where $0 < x < \frac{1}{2}$. Then by a tedious computation together with the substitution $y = g(x) > 0$, it follows that

\begin{align*} \frac{g''(x)}{g'(x)^2} = \frac{y}{y^2+4} + \frac{y}{2\sqrt{y^2+4}} - \frac{\sqrt{y^2+4}}{2y} + \frac{1}{y}. \tag{2} \end{align*}

Starting from $\text{(2)}$,

\begin{align*} \frac{g''(x)}{g'(x)^2} &= \frac{y}{\color{red}{\sqrt{y^2+4}}\cdot\sqrt{y^2+4}} + \frac{y}{2\sqrt{y^2+4}} - \color{red}{\frac{\sqrt{y^2+4}-2}{2y}} \\ &\leq \frac{y}{\color{red}{2}\sqrt{y^2+4}} + \frac{y}{2\sqrt{y^2+4}} \\ &= \frac{y}{\sqrt{y^2+4}}. \end{align*}

From this, we have the following implication:

\begin{align*} &f''(x) \geq 0, \qquad 0 < x < \tfrac{1}{2} \\[0.5em] &\impliedby \frac{y}{\sqrt{y^2+4}} \leq \frac{e^y - 1}{e^y + 1}, \qquad y > 0 \\ &\iff 1 + \frac{1}{2} y \left( y + \sqrt{y^2+4} \right) \leq e^y, \qquad y > 0 \end{align*}

Let $h(y) = 1 + \frac{1}{2} y \left( y + \sqrt{y^2+4} \right)$ denote the left-hand side. Then $h(0) = 1 = e^0$ and

$$ h'(y) = y + \frac{y^2+2}{\color{red}{\sqrt{y^2+4}}} \leq y + \frac{y^2+2}{\color{red}{2}} = 1 + y + \frac{y^2}{2} \leq e^y. $$

Therefore $h(y) \leq e^y$ as required, proving the convexity of $f(x) = \frac{1}{1+\exp(\frac{1}{x}-\frac{1}{1-x})}$ for $0 < x < \frac{1}{2}$.

Answer 2

May be an idea (?)

If you let $x=at$ the function is $$f(t)=\frac{1}{1+\exp\left(\frac{1-2 t}{t(1-t)}\right)}$$ and the derivatives are now much easier.

Expand it around $t=\frac 12$ $$f(t)=\frac{1}{2}+2 \left(t-\frac{1}{2}\right)-\frac{8}{3} \left(t-\frac{1}{2}\right)^3-\frac{416}{15} \left(t-\frac{1}{2}\right)^5+O\left(\left(t-\frac{1}{2}\right)^7 \right)$$ which is quite good for $\frac 14 \leq t \leq \frac 34$.

Removing the unnecessary terms, the last equation you wrote reduces to $$-4 t^5+14 t^4-20 t^3+16 t^2+e^{\frac{2 (2 t-1)}{(t-1) t}} \left(-4 t^5+6 t^4-4 t^3+2 t-1\right)-4 e^{\frac{2 t-1}{(t-1) t}} \left(2 t^4-5 t^3+6 t^2-4 t+1\right) t-6 t+1=0$$ The problem is to prove that $t=\frac 12$ is the only solution.