I am wondering if an inequality such as $$ \exp(x+y)\leq \exp(x) + \exp(y) $$ for certain values of $x$ and $y$. According to wolfram alpha, it should hold for the region $\mathcal{R}=\{(x,y)\in\mathbb{R}^2\colon x\leq0,y\leq 0 \}$ which for me works. My calculus is a bit rusty, but what I tried is defining the function $f(x,y)=\exp(x+y)- \exp(x) - \exp(y)$ and try to minimise/maximise it. The gradient $\nabla f(x,y)$ vanishes just at $(0,0)$, and the determinant of the Hessian matrix is negative at this point, so this is a saddle point, which makes sense since according to WA the reverse inequality holds in the complement of the region $\mathcal{R}$. How can I search for the maximum of $f$ in the desired region?
Thanks in advance.
PS: I have the feeling that I should be able to say something even stronger, but with this it would suffice.
For any $y\in\mathbb{R}$, and $x\leq 0$ then $0<\exp(y)$ and $0<\exp(x)\leq 1$ which imply that $$\exp(x+y)=\exp(x)\cdot \exp(y)\leq \exp(y) \leq \exp(x)+\exp(y).$$ By symmetry the same holds when $y\leq 0$.
P.S. If $x>0$ then it is easy to verify that $\exp(x+y)\leq \exp(x)+\exp(y)$ holds if and only if $$y\leq x-\ln(\exp(x)-1).$$ Take a look here.