Global optimum of $f(x,y)=e^{2x}(x + y^2 + 2y)$ I'm asked to find local/global optima of $f(x,y)=e^{2x}(x + y^2 + 2y)$ over $\mathbb R^2$.
Using first and second order conditions I've found that $(\frac 12,-1)$ is the only local extremum of $f$. Specifically, it's a local strict minimum. Plotting the function suggests that it also a global minimum, but how can I prove that ? I don't see any algebraic way of proving $$\forall x,y, e^{2x}(x + y^2 + 2y)\geq -\frac{e}2$$
There's a tedious way: for a fixed $y$, define $g_y(x)=e^{2x}(x + y^2 + 2y)$.
Differentiation shows that $g_y$ has a global minimum at $x=-\frac{2y^2+4y+1}{2}$ and $$g_y(-\frac{2y^2+4y+1}{2})=\frac{-e^{-2y^2-4y-1}}{2}$$
Differentiation over $y$ shows that $\displaystyle y \to \frac{-e^{-2y^2-4y-1}}{2}$ has a global minimum at $y=-1$ (the value there is $-\frac{e}2$), hence $-\frac{e}2$ is a global minimum.
My proof is very computational and would take quite some time without a computer to assist with the computations. Is there any better way ? 

Some basic manipulation of the quadratic in $y$ in this expression gives the following inequality: $$e^{2x}(x+y^2+2y)\ge e^{2x}(x-1)$$ then differentiation yields $$2e^{2x}\left(x-\frac{1}{2}\right) = 0 \Rightarrow x_{min} = \frac{1}{2}$$ thus substitution gives us $$e^{2x}(x+y^2+2y)\ge -\frac{e}{2}$$
