Prove $\ln(e^{x_1^2+x_2^2}+e^{\sqrt{x_1^2+1}})$ is convex

Question

prove $\ln(e^{x_1^2+x_2^2}+e^{\sqrt{x_1^2+1}})$ is convex over $\mathbb{R}^2$
I know that theorem

(i) Let $f : \mathbb{R}^n → \mathbb{R}$ be a convex function which is nondecreasing with respect to each of its variables separately; that is, for any i ∈ {1,2,... ,n} and fixed $x_1 x_2,..., x_{i−1}, x_{i+1},..., x_n,$ the one-dimensional function $g_i(y) = f (x_1, x_2,..., x_{i−1}, y, x_{i+1},..., x_n)$ is nondecreasing with respect to y. Let $h_1, h_2,..., h_n : \mathbb{R}^p → \mathbb{R}$ be convex functions. Prove that the composite function $r(z_1, z_2,..., z_p ) = f (h_1(z_1, z_2,..., z_p ),..., h_n(z_1, z_2,..., z_p))$ is convex.

I thought about setting $h_1(x_1,x_2)=e^{x_1^2+x_2^2},h_2(x_1,x_2)=e^{\sqrt{x_1^2+1}}$ and $g(x_1,x_2)=\ln(x_1+x_2)$ but $g$ is not convex.

any hints?

Answer 1

You can use that the sum of two log-convex functions is log-convex and apply that to $h_1$ and $h_2$.

Answer 2

We can use Mathematica with the instruction

• Plot3D[Log[E^(x^2+y^2)+E^Sqrt[x^2+1]],{x,-1,1},{y,-1,1},PlotTheme->"Detailed"]

Then we get the following picture

and this is not an answer because Martin R give a nice answer but I think it helps to see intuitively why the result about the convexity of $\displaystyle (x,y)\mapsto \log\left(e^{x^{2}+y^{2}}+e^{\sqrt{x^{2}+1}}\right)$ holds over $e^{x^{2}+y^{2}}+e^{\sqrt{x^{2}+1}}>0$ so over $e^{x^{2}+y^{2}}>-e^{\sqrt{x^{2}+1}}>0$ so over all $\mathbb{R}^{2}$.