Prove that: $\ln(e^{x+y} +1 )$
is a convex function.
I have tried to used $F \circ G$, while
$F = \ln (t+1)$ and $G = e^{x+y}$
$G$ is convex but I need to prove that $F$ is convex and growing, which I do not get. Thank you for any kind of help or hints.
Let $f(x,y)=\log(e^{x+y}+1)$. Then, we have
$$\frac{\partial f(x,y)}{\partial x}=\frac{\partial f(x,y)}{\partial y}=\frac{e^{x+y}}{e^{x+y}+1}=1-\frac{1}{e^{x+y}+1}$$
and
$$\frac{\partial^2 f(x,y)}{\partial x^2}=\frac{\partial^2 f(x,y)}{\partial y^2}=\frac{\partial^2 f(x,y)}{\partial x\partial y}=\frac{e^{x+y}}{(e^{x+y}+1)^2}>0$$
Therefore, since the non-mixed second partial derivatives are positive (we only need non-negative) and the determinant of the Hessian is zero (again, we only need it to be non-negative), then we conclude that $f$ is convex.