Lambert W function I wan to show that, for $x \in [-\frac{1}{\exp(1)},0)$, $W_0(x) + W_{-1}(x) \leq -2$. Can any body help or give a suggestion? It seems trivial from the graph of functions but I need a rigorous proof. 
Thanks in advance,
Reza
 A: Let $a=W_{-1}(x)$ and $b=W_0(x)$ so the desired inequality is $a+b \le -2$. Now note that if we already have $a \le -2$ we are done, so that we may assume in fact $-2<a\le -1.$ 
Then $a+b\le -2$ will hold provided $b \le -2-a$. The idea now is to show that, for each $a \in [-2,-1],$ the point $-2-a$ lies to the right of $b$. If this is shown, then the value of $a+b$ will be at most $-2$ because $h(x)=xe^x$ is increasing on $[-1,0]$ and so $-2-a$ has to be decreased to become $b$. [One has to think aabout this to get it straight, and maybe I could explain it better...]
Anyway if we can show that $h(-2-a) \ge h(a)$ on $[-2,-1]$ we are in business as outlined above. Define $p(t)=h(-2-t)-h(t),$ [recall $h(x)=xe^x$], and then we want to show $p\ge 0$ on $[-2,-1]$. We have $p(-2)=2e^{-2}>0$ and $p(-1)=0$. A graph shows $p$ is a nice decreasing curve on $[-2,-1]$ which then would imply $p \ge 0$ as desired. However one wants a rigorous proof, so naturally calculates $p'(t)$ and luckily it factors nicely:
$$p'(t)=(t+1)\cdot [ e^{-2-t}-e^t].$$
The only zero of the first factor is at $t=-1$, and some manipulations show the second factor is also zero only at $t=-1.$ These are endpoints of the interval $[-2,-1]$ which we're interested in, so that we now have shown that $p$ has no interior critical points on this interval. It follows that $p$ is decreasing as desired, and since it is $0$ at the right endpoint $-1$ we have in fact that $p(t)\ge 0$ to finish the proof.
A: I found a simpler solution.
Simple algebraic manipulation shows that the parabola $e^{-1}(\frac{1}{2} x^2 + x -\frac{1}{2})$ is below the function $x e^x$ for $x \in (−1, 0)$ and above it for $x < −1$. Therefore, any line parallel to the $x$-axis with the $y$-intercept in the interval $y = [−\frac{1}{e}, 0)$, intersects the parabola at equal or larger $x$ coordinates than those corresponding to intersections with $xe^x$. From the
other side, the $x$ coordinate of the intersection points of these lines with $x e^x $ gives $W_0(y)$ and $W_{−1}(y)$. Consequently, for any $y \in [−\frac{1}{e}, 0)$, the sum $W_0(y) + W_{−1}(y)$ is less than or equal to the sum of the roots of $e^{-1} (\frac{1}{2} x^2 + x -\frac{1}{2}) = y$ which is always equal to $−2$. This completes the proof.
