Find $x$ that minimizes $\max_{1 \leq i \leq 3} f_i(x)$ My question is:

For $0 < a_1 \leq a_2 \leq a_3$ fixed, find $x>0$ that minimizes
\begin{align*}
g(x) = \max_{1\leq i \leq 3} f_i(x).
\end{align*}
  where
  \begin{align*}
f_i(x) =  \left(\frac{a_i - x}{a_i + x}\right)^2
\end{align*}

I do not know how to start. Any idea, reference is welcome. Thank you!
 A: I think I have a solution to my question.

Answer: the minimizer is $x = \sqrt{a_1 a_3}$.

Let $0 < \underline{a} < \overline{a}$ be fixed. We study the relative position of $f_{\underline{a}}$ and $f_{\overline{a}}$ defined by
\begin{align*}
f_{\underline{a}}(x) = \left(\frac{\underline{a} - x}{\underline{a} + x}\right)^2, \qquad 
f_{\overline{a}}(x) = \left(\frac{\overline{a} - x}{\overline{a} + x}\right)^2
\end{align*}
for $x>0$. Let $h$ be defined by $h = f_{\underline{a}} - f_{\overline{a}}$. Then,
\begin{align*}
h(x) &= \frac{(\underline{a}-x)^2(\overline{a} + x)^2 - (\overline{a} - x)^2(\underline{a}+x)^2}{(\underline{a}+x)^2(\overline{a}+x)^2}\\
&= \frac{\big[(\underline{a}-x)(\overline{a} + x) - (\overline{a} - x)(\underline{a}+x)\big]\big[(\underline{a}-x)(\overline{a} + x) + (\overline{a} - x)(\underline{a}+x)\big]}{(\underline{a}+x)^2(\overline{a}+x)^2}\\
&= \frac{4 x (\overline{a} - \underline{a})(x^2 - \underline{a}\overline{a})}{(\underline{a}+x)^2(\overline{a}+x)^2}.
\end{align*}
Hence, $f_{\underline{a}} < f_{\overline{a}}$ for $x \in (0, \sqrt{\underline{a}\overline{a}})$, $f_{\underline{a}} = f_{\overline{a}}$ for $x = \sqrt{\underline{a}\overline{a}}$, and $f_{\underline{a}} > f_{\overline{a}}$ for $x \in (\sqrt{\underline{a}\overline{a}}, +\infty)$. Since $0 < a_1 < a_2 < a_3$, we deduce that
\begin{align*}
&f_1 \leq f_2 \quad \text{if} \quad x \leq \sqrt{a_1 a_2}, \quad\text{otherwise} \quad f_1>f_2,\\
&f_1 \leq f_3 \quad \text{if} \quad x \leq \sqrt{a_1 a_3}, \quad\text{otherwise} \quad f_1>f_3, \\
&f_2 \leq f_3 \quad \text{if} \quad x \leq \sqrt{a_2 a_3}, \quad\text{otherwise} \quad f_2 > f_3.
\end{align*}
Consequently, the maximum is given by
\begin{align*}
g(x) = \begin{cases}
f_3(x) & \text{if }x \leq \sqrt{a_1 a_3}\\
f_1(x) & \text{if }x > \sqrt{a_1 a_3}.
\end{cases}
\end{align*}
On the other hand, for $i \in \{1, 2, 3\}$, 
\begin{align*}
f_i'(x) = \frac{4 a_i (x- a_i)}{(a_i+x)^3}.
\end{align*}
Hence, $f_i$ is decreasing on $(0, a_i]$, and increasing on $(a_i, + \infty)$. This implies that $f_3$ is decreasing on $(0, \sqrt{a_1 a_3}]$, and $f_1$ is increasing on $(\sqrt{a_1a_3}, + \infty)$. Hence, $g(x)$ is the smallest for $x = \sqrt{a_1 a_3}$.
A: You can introduce a new variable $z$ and minimize $z$ subject to $$z \ge \left(\frac{a_i-x}{a_i+x}\right)^2$$ for all $i$.  This reformulation avoids the nondifferentiable $\max$ function.  Maybe you can try the method of Lagrange multipliers.
