inverse of function $xe^x$

Question

Let W the Lambert function that is the inverse function of $xe^x$.Is true that if $c > 1$, then $w(cx)\leq cw(x)$ for $x\geq 0$?

Can I apply the property $ln(x)-ln(ln(x))\leq W$ for $x\geq e$?

Answer 1

Step 1. We want to prove that for any $x\geq 0$ and $c>1$ we have $$ W(cx)\leq c\cdot W(x)$$ where $W$ is Lambert's function.

Step 2. $xe^x$ is an increasing function on $\mathbb{R}^+$. It follows that if $a$ and $b$ are positive numbers, proving $a\leq b$ is equivalent to proving $a e^a\leq b e^b$. We apply this principle to the above inequality, turning it into

$$ W(cx) e^{W(cx)} \leq c W(x) e^{c W(x)}. $$

Step 3. By the very definition of $W$, the LHS of the last inequality equals $cx$, while the RHS equals $cW(x) e^{W(x)} e^{(c-1)W(x)}$. Thus the last inequality is equivalent to

$$ cx \leq cx e^{(c-1)W(x)} $$ or to $$ 1 \leq e^{(c-1)W(x)} $$ which is trivial.