Prove that $W(x) \geq (\ln(x) + 1)/2$ for $x \geq 1$, where $W$ is the Lambert-W function.

Question

I saw this used in a proof in a paper. What is the simplest proof? Does it use $W(x)e^{W(x)} = x$ directly, or do we have to use the known lower bound $\ln x - \ln \ln x + \frac{\ln\ln x}{2\ln x} \leq W(x)$? (https://en.wikipedia.org/wiki/Lambert_W_function#Asymptotic_expansions)

Answer 1

Consider the function $$f(x)=W(x)-\frac{1}{2} (\log (x)+1)$$ You have $$f'(x)=\frac{W(x)}{x (W(x)+1)}-\frac{1}{2 x}$$ which cancels at $x=e$ and $f(e)=0$. The second derivative $f''(e)=\frac{1}{8 e^2}$ shows that this is a minimum.