Composition of a convex function and a convex decreasing function is quasi-concave

Let $h$ be a convex decreasing function and $g$ a convex function. Is it true that $h(g(x))$ is a quasi-concave function?

Yes.

Since $g$ is convex, for $\lambda \in [0,1],$

$$g(\lambda x +(1-\lambda )y) \leq \lambda g(x) + (1-\lambda)g(y).$$

and

$$\lambda g(x) + (1-\lambda)g(y)\leq\max[g(x),g(y)]$$

Since h is decreasing,

$$h[g(\lambda x +(1-\lambda )y)] \geq h[\lambda g(x) + (1-\lambda)g(y)]\geq h(\max[g(x),g(y)])\geq \min[h(g(x)),h(g(y))].$$

You don't need $h$ to be convex. Also, you only need $g$ to be quasiconvex. The following two facts should be observed:

1. Composition with any increasing function preserves quasi-convexity and quasi-concavity
2. A function $f$ is quasiconvex if and only if $-f$ is quasiconcave.
3. Therefore, composition with any decreasing function turns quasi-convexity into quasi-concavity, and the other way around.