Prove function in two variables is convex

Question

Prove that $$f:(x,y)\to\log(\exp(x)+\exp(y))$$ on $\mathbb{R}^2$ is convex.

By direct computation, the Hessian matrix is $H\begin{bmatrix}a&-a\\-a&a\end{bmatrix}$ where $a=\tfrac{e^{x+y}}{(e^x+e^y)^2}>0$. The eigenvalues of $H$ are $0$ and $2a$, both of which are nonnegative, so $H$ is positive semidefinite, so $f$ is convex.

Can you solve the problem using the $0\le\lambda\le 1$ definition of convexity? I got that it was equivalent to $a_1^\lambda b_1^{1-\lambda}+a_2^{\lambda}b_2^{1-\lambda}\le(a_1+a_2)^\lambda(b_1+b_2)^{1-\lambda}$ for $0\le\lambda\le 1$ and $a_1,a_2,b_1,b_2>0$. I tried to combine different weighted AM-GM inequalities/repeatedly use a 1D inequality to get the 2D inequality, but I was unsuccessful.

Answer 1

If $0\leq t \leq 1$ then $t^{\lambda} s^{1-\lambda}+(1-t)^{\lambda} (1-s)^{1-\lambda} \leq 1$. To see this differentiate w.r.t. $t$ to see that the maximum is attained when $\frac t {1-t}= \frac s {1-s}$ or $t=s$ (in which case equality holds). Now take $t=\frac {a_1} {a_1+a_2}$ and $s=\frac {b_1} {b_1+b_2}$

Answer 2

Another way : Consider a level set $S=f^{-1}(t)$

Here out unit normal $N$ (cf gradient) is a multiple of $ (e^x,e^y)$. Then $(x,y)\neq (a,b)\in S$, assume that $(e^x,e^y)=C(e^a,e^b)$ where $e^x+e^y=e^a + e^b$. Hence we have $C=1$.