On branches of Lambert W function

Question

When I solve the inequality

$$e^{\sqrt x} \geq 2x $$

in Wolfram Alpha, I get the solution

$$0\leq x \leq 4 W(-\frac{1}{2\sqrt 2})^2 $$

and

$$x\geq 4 W_{-1}(-\frac{1}{2\sqrt 2})^2 .$$

My questions:

-How did get this results?

-Why there exist two solutions of this inequality? What is the difference between $W(x)$, $W_{0}(x)$ and $W_{-1}(x)$? (This question more important then the other.)

Answer 1

Obviously, $x\ge 0$ for the square root. Take the square root on both sides, $$ e^{\sqrt x/2}\ge 2\sqrt2(\sqrt x/2)\iff -\frac1{2\sqrt2}\le(-\sqrt x/2)e^{-\sqrt x/2} $$ Now the function $ue^u$ has

• a negative minimum at $u=-1$ with value$-e^{-1}$,
• value $ue^u=0$ at $u=0$ and
• $\lim_{u\to\-\infty}ue^u=0$.

Thus for $-e^{-1}<v<0$ the equation $v=ue^u$ has two solutions $W_{-1}(v)<-1$ and $-1<W_0(v)<0$, which are on the two branches of the Lambert-W function.

• $W_{-1}$ is the inverse of the monotonically falling segment and thus also monotonically falling from $-1$ to $-\infty$,
• for the same reasoning $W_0$ is monotonically increasing from $-1$ to $0$.

As $2\sqrt2>e$, if barely, you get these two interval end points on the branches and all $x$ to the left and right as solution of the inequality, $$ -\sqrt x/2\ge W_0\left(-\frac1{2\sqrt2}\right)\text{ and }W_{-1}\left(-\frac1{2\sqrt2}\right)\ge -\sqrt x/2 $$