While working on a variational problem I have reached to the following question:

Let $f:(0,\infty) \to [0,\infty)$ be a $C^1$ function satisfying $f(1)=0$. Suppose that $f(x)$ is strictly increasing on $[1,\infty)$ and strictly decreasing on $(0,1]$. Define $F:(0,1) \to [0,\infty)$ by $$ F(s)=\min_{xy=s,x,y\in(0,\infty)} f(x)+ f(y). $$

**Question:** For which functions $f$, $F(s)$ has an affine part? Can we
characterize such functions?

The motivation is that I am applying Jensen inequality with $F$, and an affine part (in contrast to *strict* convexity) gives some flexiblity.

The only example that I know of is when $f(x)=(x-1)^2$, and $$ F(s) = \begin{cases} 1-2s, & \text{ if }\, 0 \le s \le \frac{1}{4} \\ 2(\sqrt{s}-1)^2, & \text{ if }\, s \ge \frac{1}{4}, \end{cases} $$ is affine on $[0,\frac{1}{4}]$.

**Is this $f$ the only choice which makes $F$ affine?**

For qubic and quartic penalizations this is not the case; if $f(x)=(1-x)^3$, then $$ F(s)=\begin{cases} 1 - 3 s - 2s^{3/2} &\text{ if } 0<s\le1/9, \\ 2 + 6 s - 2(3 + s)s^{1/2} &\text{ if } 1/9\le s<1. \end{cases} $$ and similarly for $f(x)=(x-1)^4$.

*Here is an attempted analysis:*

Assume that there is a $C^1$ map $s \to (x(s),y(s))$ giving a minimizer to the problem, i.e. for any $s \in (0,1]$ $$ F(s)=f(x(s))+f(y(s)), \, \, \,x(s)y(s)=s. \tag{1} $$ Lagrange's multipliers give $$ f'(x(s))=\lambda(s) y(s),f'(y(s))=\lambda(s) x(s). \tag{2} $$ $F'(s)=\lambda (s)$, so $F''(s)=0$ if and only if $\lambda(s) < 0$ is constant. ($\lambda < 0$ since $f'|_{(0,1)} < 0$ by our assumption.)

*I don't see how to proceed from here.*

For $f(x)=(x-1)^2$ we have $\lambda(s)=-2$: The minimum (for $s \in [0,\frac{1}{4}]$) is obtained at $x(s)+y(s)=1$, so $ f'(x(s))=2(x(s)-1)=-2y(s). $