Prove for strictly decreasing $f$'s, and $x_n \to 1$ the limit of the average of the $f'_n(\cdot)$'s on $(x_n,1]$ divided by $f_n'(x_n)$ is $<1$

Question

I have a sequence of smooth functions $f_n : [ 0, 1]\rightarrow [0,1]$ with $f_n^\prime(\cdot)<0$.

Assume that there is a pair of sequences, $x_n \rightarrow 1$ and $\delta_n \rightarrow 0$ such that for each $n$, $\delta_n f_n^\prime (x_n ) < -1$.

Let $\mu_n = \frac{\int_{x_n}^1 f_n^\prime(\chi)d\chi}{1-x_n}$. That is, $\mu_n$ is the average of the $f_n^\prime$ values between $x_n$ and 1.

I want to show that $\lim_{n\rightarrow \infty} \frac{\mu_n}{f_n(x_n)} <1$.

It seems obvious that this must be true, since for each $n$, $f_n(\cdot) < f_n(x_n)$ on $(x_n, 1]$, and there is an infinite number of points between $x_n$ and 1 at which $f_n^\prime(\cdot)$ is very negative. But except for continuity, I don't know anything about the rest of the points between $x_n$ and 1.