Showing that the function $f(x,y)=x+y-ye^x$ is non-negative in the region $x+y≤1,x≥0,y≥0$ ok, since it's been so long when I took Calculus, I just wanna make sure I'm not doing anything wrong here. 
Given $f:\mathbb{R}^2\rightarrow \mathbb{R}$ defined as $f(x,y)=x+y-ye^x$. I would like to show that the function is nonnegative in the region $x+y\leq 1, \;\;x\geq 0, \;\;y\geq 0$. 
Now my game plan is as follows:
1. Show the function is non-negative on the boundary of the region
2. Show the function takes a positive value in the interior of the region
3. Show that the function has no critical points in the interior of the region
4. By continuity the function is non-negative everywhere in the region. 
Is the above sufficient or am I doing something wrong? Would there be a better way to show this? 

I'll try using Lagrange multiplier:
The function is:
$$f(x,y) = x + y - ye^x$$
and constraint are:
$$g(x,y) = x+y \leq 1$$
$$h(x) = x \geq 0$$
$$j(y) = y \geq 0$$
So using Lagrange multiplier now we have:
$$F(x,y,\lambda,\lambda_1,\lambda_2) = x + y - ye^x - \lambda(x+y-1) - \lambda_1(x) - \lambda_2(y)$$
Now we take parital derivatives:
$$F_x = 1 - ye^x - \lambda - \lambda_1 = 0$$
$$F_y = 1 - e^x - \lambda - \lambda_2 = 0$$
$$\lambda(x+y-1) = 0$$
$$\lambda_1(x) = 0$$
$$\lambda_2(y) = 0$$
Now we have 8 cases:
1) $\lambda = \lambda_1 = \lambda_2 = 0$
This implies one solution $(x,y) = (0,1)$
2) $\lambda = \lambda_1 = y = 0$
Now in $F_x$ we have $1=0$, which is not posible, so this case doesn't give a solution.
3) $\lambda = x = \lambda_2 = 0$
Now in $F_x$ we have $y + \lambda_1 = 1$, because all $\lambda$ values are positive, we get $y \leq 1$. So the solutions are $(x,y) = (0,y)$, where $0 \leq y \leq 1$
4) $\lambda = x = y = 0$
Simply this implies one solution $(x,y) = (0,0)$
5) $x + y - 1 = \lambda_1 = \lambda_2 = 0$
This implies a solution that we've already obtained $(x,y) = (0,1)$
6) $x + y - 1 = \lambda_1 = y = 0$
This simply implies one solution $(x,y) = (1,0)$
7) $x + y - 1 = x = \lambda_2  = 0$
This simply implies one solution $(x,y) = (0,1)$
8) $x + y - 1 = x = y = 0$
This case doesn't imply any solution because it's a contradiction.
Now we have 4 distinct solutions we check them all now:
1) $(x,y) = (0,1)$
$$f(x,y) = x + y - ye^x = 1 - 1 = 0$$
2) $(x,y) = (0,y)$
$$f(x,y) = x + y - ye^x = y - y = 0$$
3) $(x,y) = (0,0)$
$$f(x,y) = x + y - ye^x = 0$$
4) $(x,y) = (1,0)$
$$f(x,y) = x + y - ye^x = 1$$
This means that $f(x,y)$ has minimum of $0$ at point $(0,y)$ and maximum of $1$ at point $(1,0)$.
Q.E.D.
