Prove that if $\alpha \in BV[a,b]$ and for some positive number $M$ we have $|\alpha (x)| > M$ for $x$ in $[a,b]$, then $1/\alpha \in BV[a,b]$.
I am trying to use the definition of bounded variation with the mean value theorem but I'm stuck. What do I do with the inequality?
We have that $|\alpha(x)| > M$, so that the reciprocal is well defined, at the least. Also, since $\alpha$ is of bounded variation, we'll let the constant $C$ be the variation of $\alpha$ (the constant that bounds the sum you gave, for every possible partition).
Next, consider any partition of $[a,b]$, say $x_i$, $0 \leq i \leq n$, where $x_0 = a$ and $x_n = b$.
Now, we basically have to estimate the quantity: $$ \sum_{i=0}^{n-1} |f(x_{i+1}) - f(x_{i})| = \sum_{i=1}^n \left|\frac{1}{\alpha(x_{i+1})} - \frac{1}{\alpha(x_i)} \right| $$
Expand the inner bracket: $$ \sum_{i=0}^{n-1} \left|\frac{1}{\alpha(x_{i+1})} - \frac{1}{\alpha(x_i)} \right| = \sum_{i=0}^{n-1} \left|\frac{\alpha(x_i) - \alpha(x_{i+1})}{\alpha(x_i) \alpha(x_{i+1})}\right| $$
Now, since $|\alpha(x)| > M$, it follows that $\frac{1}{|\alpha(x)|} < \frac 1M$ for all $x \in [a,b]$. Hence: $$ \left|\frac{\alpha(x_i) - \alpha(x_{i+1})}{\alpha(x_i) \alpha(x_{i+1})}\right| \leq |\alpha(x_i) - \alpha(x_{i+1})| \frac{1}{|\alpha(x_i)|} \frac{1}{|\alpha(x_{i+1})|} \leq \frac{|\alpha(x_i) - \alpha(x_{i+1})|}{M^2} $$
Summing on both sides: $$ \sum_{i=0}^{n-1} \left|\frac{\alpha(x_i) - \alpha(x_{i+1})}{\alpha(x_i) \alpha(x_{i+1})}\right| \leq \frac 1{M^2}\sum_{i=0}^{n-1} |\alpha(x_i) - \alpha(x_{i+1})| \leq \frac{C}{M^2} $$
Hence, you get that $\frac 1 \alpha$ is also of bounded variation, since every possible variation is bounded by a finite quantity $\frac C{M^2}$.