Exercise Suppose that $\{f_n\}$ converges uniformly to $f$ on an interval $I$ and that there is a positive constant $M$ such that $f_n(x) \geq M$ for each $n \in \mathbb{N}$ and $x \in I$. Show that $\{\frac{1}{f_n}\}$ converges uniformly to $\frac{1}{f}$ on $I$.
Proof Attempt
Since $f_n(x) \geq M \hspace{0.2cm} \forall n \in \mathbb{N}, \forall x \in I$, we have that
$$\frac{1}{f_n(x)} \leq \frac{1}{M}$$
Consider the constant sequence $ \langle C_1, C_2, C_3, \dots \rangle = \langle \frac{1}{M}, \frac{1}{M}, \dots \rangle$
By our hypothesis, for each constant $C_k \in \{C_i\}_{i=1}^\infty$, we have
$$\forall n \in \mathbb{N}, \, \forall x \in I, \hspace{0.2cm} \frac{1}{f_n(x)} \leq C_k$$
To me, it would seem beneficial to use the Weierstrass $M$-Test here so that we could conclude that the $\frac{1}{f_n}$'s are uniformly Cauchy, and thus uniformly convergent. The only glaring issue is that there is no way to guarantee that $\displaystyle\sum_{k=1}^\infty C_k < \infty$. Am I taking this proof in the right direction? Is the $M$-test applicable here?
For uniform convergence we have equivalent condition $$\lim\limits_{n\to\infty}\sup\limits_I|f_n(x)-f(x)|=0$$ Now considering $$\left|\frac{1}{f_n}- \frac{1}{f_n} \right|=\left|\frac{f_n-f}{f_nf} \right|$$ we have needed estimation, because $f\geqslant M$ together with $f_n$.