Is the optimum of this problem convex in the constraint parameter? Let  $f:\mathbb R^+ \to \mathbb R$ be a smooth function, satisfying $f(1)=0$, and suppose that 
$|f|$ grows with the distance from $1$: $|f(x)|$ is strictly increasing when $x \ge 1$, and strictly decreasing when $x \le 1$.
Suppose also that $\lim_{x \to \infty} |f(x)| = \infty$.
For any $s \in (0,1)$, define
$$
F(s)=\min_{xy=s,x,y>0} f^2(x)+ f^2(y) 
$$
(The minimum exists since $|f|$ diverges at infinity.)

Question: Does there exist a convex function $g(s)$ such that $F=g^p$ for some $p \ge 1$? I do not require $g$ to be positive.

Here are two examples where this happens:
Linear penalization: $f(x)=x-1$. In that case
$$F(s) =
\begin{cases}
2(\sqrt{s}-1)^2,  & \text{ if  }\, s \ge \frac{1}{4} \\
1-2s, & \text{ if  }\, s \le \frac{1}{4}
\end{cases}
$$
is convex, since $F'(s)$ is non-decreasing.
Logarithmic penalization: $f(x)=\log x$. In that case
$$ F(s)=2f^2(\sqrt s)=\frac{1}{2}(\log s)^2$$ is not convex.
However, we have $F(s)=g^2(s)$ where $g(s)=-\frac{1}{\sqrt 2}\log s$ which is convex.
Is there a general phenomena lying behind these two examples?
 A: The answer is no. E.g., let $f(x):=|x-1|^{3/2}$. Then 
$$
F(s)=\begin{cases}
 F_1(s) &\text{ if } 0<s\le1/9, \\
 F_2(s) &\text{ if } 1/9\le s<1,
\end{cases}
$$
where 
$$F_1(s):=1 - 3 s - 2s^{3/2},$$
$$F_2(s):=2 + 6 s - 2(3 + s)s^{1/2}.$$
One may note here that $F_1(1/9)=16/27=F_2(1/9)$ and $F'_1(1/9)=-4=F'_2(1/9)$.
From the definition of $F$, it is clear that $F>0$ on $(0,1)$. So, letting $a:=1/p\in(0,1]$, we see that the desired goal was that 
$h:=F^a$
be either convex or (if $p$ is even) concave. 
(If $h$ is concave and $p$ is even, we can take $g:=-h$. Then $g$ will be convex and we will also have $g^p=h^p=F$.)
So, letting $h_j:=F_j^a$ for $j=1,2$, we see that we must have one of the following cases: 
(i) $h_1$ is convex on $(0,1/9]$ and $h_2$ is convex on $[1/9,1)$; 
(ii) $h_1$ is concave on $(0,1/9]$ and $h_2$ is concave on $[1/9,1)$. 
However, $h_1''(1/9)$ equals $3a-4$ in sign and hence is $<0$ for $a\in(0,1]$, whereas $h_2''(1/9)$ equals $2+3a$ in sign and hence is $>0$ for $a\in(0,1]$. So, neither one of the cases (i) or (ii) can take place. 
Here is the graph of $F''$: 

