Convexity of $h=g+\eta\sqrt{f-g^2}$

Question

Suppose for all $\eta\in[0,1]$ and $\gamma>0$, we define $h:R^{\geq0}\to R^{\geq 0}$ in terms of two convex functions as follows $$h=g+\eta\sqrt{f-g^2}$$ where $f:R^{\geq0}\to R^{\geq 0}$ and $g:R^{\geq0}\to R^{\geq 0}$. We also know that $f,g\geq0$, $f',g'\leq0$, $f'',g''\geq0$, and $f'=-\gamma g$. How can we show the convexity of $h$?

Answer 1

The conjecture is false. Just take $\eta=\gamma=1$, $f(x)=g(x)=e^{-x}$. In this case $h(x)$ is concave.