Consider $f:\mathbb{R}^{3}\rightarrow\mathbb{R}$ such that $(x,y)\mapsto \ln(e^{x}+e^y+e^z)$. Determine whether or not $f$ is convex and rigorously justify your answer.
I am thinking this function is not convex just based on intuition.
I have taken second order partial derivative and get $\nabla^2=\langle \dfrac{e^x+e^y}{e^x+e^y+e^z},\dfrac{e^y+e^z}{e^x+e^y+e^z},0\rangle$,
So, I believe I want to show the gradient is greater than or equal to zero to show convex,
Am I on the right path and if so what should my next step be.
Thank you
If you calculate the Hessian matrix, you should get something like
$$ \frac{1}{\left(e^x+e^y+e^z\right)^2} \left( \begin{array}{ccc} e^x \left(e^y+e^z\right) & -e^{x+y} & -e^{x+z} \\ -e^{x+y} & e^y \left(e^x+e^z\right) & -e^{y+z} \\ -e^{x+z} & -e^{y+z} & e^z \left(e^x+e^y\right) \\ \end{array} \right) $$
Now, in order to hint at $f$ being convex, one needs to show that this Hessian is positive semi-definite, but to this end, it is sufficient to notice that the Hessian matrix is symmetric, has non-negative diagonal entries, and is diagonally dominant.