Why inverse Mills ratio for normal distribution is 1-Lipschitz continuous?

Question

The inverse Mill ratio for a standard normal distribution is: $$ IMR(x) = \frac{\phi(x)}{\Phi(x)}, $$ where $\phi(x)$ is the pdf of standard normal distribution and $\Phi(x)$ is the cdf of standard normal distribution. The paper "Nonasymptotic analysis of semiparametric regression models with high-dimensional parametric coefficients" by Ying Zhu suggests that this function is 1-Liphshits continuous (see note at page 2290): $$ |IMR(x) - IMR(y)| \leq |x - y|. $$

How one should rigorously prove this statement?

Answer 1

The inverse Mill ratio for a standard normal distribution is usually denoted by $\lambda(x)$ and verifies $$\lambda'(x) \in (-1,0) \quad \forall x.$$

See https://math.stackexchange.com/q/1367026 for a proof of $\lambda'(x)<0$ and https://math.stackexchange.com/q/3118157 for a proof of $\lambda'(x)>-1$.