Ratio of CDF to PDF increasing?
The inverse Mill's ratio is defined as $$ \lambda(x)=\frac{\phi(x)}{\Phi(x)}. $$ For answering the question, it suffices to show that $\lambda'(x)<0$. Note that the p.d.f. of standard normal $\phi$ is differentiable. Thus we can apply quotient rule to it $$ \lambda'(x)=\frac{\phi'(x)\Phi(x)-\phi(x)^2}{\Phi(x)^2}=\frac{-x\phi(x)\Phi(x)-\phi(x)^2}{\Phi(x)^2}=-\lambda(x)(x+\lambda(x)). $$ Observe that $$ \phi'(x)=-x\frac{1}{\sqrt{2\pi}}\exp\left\lbrace -\frac{x^2}{2}\right\rbrace=-x\phi(x). $$ It is clear that $$ x+\lambda(x)>0,\quad \forall x\geq 0. $$ The challenge is to show that $$ -x+\lambda(-x)>0,\quad \forall x>0. $$ Here we exploit two facts:
normal distribution is symmetrical around $0$.
the inverse Mill's ratio is the expectation of the truncated normal $$ \frac{\phi(x)}{1-\Phi(x)}=E[X|X>x],\quad X\sim \mathcal{N}(0,1). $$
Thus the condition we want to show can be written as $$ E[X|X>x]=\frac{\phi(x)}{1-\Phi(x)}=\frac{\phi(-x)}{\Phi(-x)}=\lambda(-x)>x,\quad \forall x>0. $$ If this is not immediate to you, invoke Chebychev's inequality to get $$ E[X|X>x]\geq x. $$ For strict inequality, argue by contradiction. Suppose $E[X|X>x]=x$, then it is necessary that $$ \int_{x}^\infty (X-x)d\Phi(X)=0, $$ which implies that $X=x$ except on a set of $\Phi$ measure zero. But this is clearly false that we can always find a set with strictly positive $\Phi$ measure for which $X>x$ given any $x$.