$f$ is concave, $g:\Bbb R \to \Bbb R$ is decreasing, prove that $g\circ f$ is convex

Question

Please prove that if the function $f$ is concave and $g:\Bbb R \to \Bbb R$ is decreasing, then $g\circ f$ is convex.

Answer 1

Take $s,t \geq 0$ with $s + t = 1$. Take $x,y \in \Bbb R$. By definition of concavity, $$ f(sx + ty) \geq sf(x) + tf(y) $$ By definition of a decreasing function, this means that $$ g(f(sx + ty)) \leq g(sf(x) + tf(y)) $$ At this point, we would need an additional statement like "$g$ is convex" to state that $$ g(sf(x) + tf(y)) \leq s\,g(f(x)) + t\,g(f(y)) $$ however, outside of that, I don't believe the statement holds.

Answer 2

The identity function $f(x) = x$ is concave, so the result implies every decreasing function is convex. Doubtful...