Is $f^{-1}\big( \sqrt{xf(x)} \big)$ convex (for large $x$) when $f(x) = o(x)$ is concave and strictly increasing? Question
Let $f : (N, \infty) \to \mathbb (0,\infty)$ be a strictly increasing, concave function such that $f(x) \to \infty$ and $f(x) = o(x)$ for $x \to \infty$, where $N > 0$. Also, let $f$ be twice differentiable (or more if needed).
Define $g(x):= f^{-1}\big(\sqrt{xf(x)} \big)$ where the definition makes sense (i.e. for all $x$ such that $\sqrt{x f(x)} > f(N)$).
Does it follow that $g$ is convex for large $x$? That is, if the restriction $g|_{[K, \infty)}$ is convex for some $K>0$.
Also: I'd like to know if $h(x):= g(x)/x$ is increasing for large $x$.

Thoughts
The idea is that the geometric mean $\sqrt{x f(x)}$ sort of pulls $f$ up toward the 45 degree line, straighening it out a bit. This is still concave (which isn't too hard to show, knowing the geometric mean is a concave function), but "less" so. Applying the convex $f^{-1}$ to $f(x)$ yields $x$, sort of cancelling out the concavity. But applying $f^{-1}$ to $[xf(x)]^{1/2}$ should yield a convex function, since we already brought $f(x)$ a bit in this direction with the geometric mean.
 A: Computing the second derivative of $g$ shows that convexity of $g$ at $x$ is equivalent to
$$
\frac{y''(x)}{y'({x})} \geq \frac{f''(\tilde{x})}{f'(\tilde{x})}
$$
where $y(x) = \sqrt{xf(x)}$ and $\tilde{x}$ is defined by
$$
f(\tilde{x})= y(x),
$$
which formalizes your strategy to proceed via showing that "$y$ is more convex than $f$". I wasn't able to show this inequality.
A partial answer to your second question:

For any sufficiently smooth function $g$, if $g$ is eventually uniformly convex, then $g(x)/x$ is eventually increasing.
(Uniform convexity of $f$ on $(a, b)$ means that there exists a constant $C>0$ such that $f''(x)>C$ for all $x\in (a, b)$.)

Proof: Assume $g$ is uniformly convex on $(x_0, \infty)$. Then we have
$$
g(x) = g(x_0) + \int_{x_0}^{x} g'(y) dy \\
= g(x_0) + \int_{x_0}^x (g'(x) - \int_{y}^x g''(z)dz) dy \\
\leq g(x_0) + (x-x_0) g'(x) - \frac{(x-x_0)^2}{2}C
$$
for all $x>x_0$,
which we may rewrite as $\frac{g(x)-g(x_0)}{x-x_0} + \frac{x-x_0}{2}C \leq  g'(x)$. If we pick $x_1>x_0$ such that $\frac{x_1-x_0}{2}C\geq \frac{g(x_0)}{x_1-x_0}$, we obtain
$$
g'(x) \geq g(x)/x
$$
for all $x>x_1$ and therefore
$$
\frac{d}{dx} \frac{g(x)}{x} = \frac{g'(x)x - g(x)}{x^2} > 0
$$
for all $x>x_1$.
And

There is a strictly convex function $g(x)$ such that $g(x)/x$ is strictly decreasing

Proof: $g(x):=\sqrt{1+x^2}$.
