Suppose that for $f,g$ of bounded variation and $c \in \mathbb{R}$ that $cf, f + g, fg$ are all of bounded variation.
Using these, I'd like to show that $\max\{f,g\}$ is of bounded variation too, but am having some trouble towards the end.
Recall the identity,
$$\max\{x,y\} = \frac{1}{2}(x + y + |x - y|).$$
We then have $$\sum_n\left|\frac{1}{2}\left(f(x_i) + g(x_i) + |f(x_i) - g(x_i)|\right) - \frac{1}{2}\left(f(x_{i-1}) + g(x_{i-1}) + |f(x_{i-1}) - g(x_{i-1})|\right)\right|$$
Since $|cf - cg| = |c||f-g|$, and doing some rearranging
$$\frac{1}{2}\sum_n\left|f(x_i) - f(x_{i-1})+ g(x_i) - g(x_{i-1}) + |f(x_i) - g(x_i)| - |f(x_{i-1}) - g(x_{i-1})|\right| \\ \stackrel{\text{T.I.}}{\leq} \frac{1}{2}\sum_n\left|f(x_i) - f(x_{i-1})+ g(x_i) - g(x_{i-1})\right| + \frac{1}{2}\sum_n\left||f(x_i) - g(x_i)| - |f(x_{i-1}) - g(x_{i-1})|\right| $$
For the first sum
$ \begin{array}{c} \frac{1}{2}\sum_n\left|f(x_i) - f(x_{i-1})+ g(x_i) - g(x_{i-1})\right| &\stackrel{\text{T.I.}}{\leq} \frac{1}{2}\sum_n\left|f(x_i) - f(x_{i-1})\right| + \frac{1}{2}\sum_n\left|g(x_i) - g(x_{i-1})\right| \\ &= \frac{1}{2}V_f + \frac{1}{2}V_g \hspace{6.4cm}\\ \end{array} $
For the right I am not sure (reverse triangle inequality possibly?): $$\frac{1}{2}\sum_n\left||f(x_i) - g(x_i)| - |f(x_{i-1}) - g(x_{i-1})|\right|$$
By reverse triangle inequality $$| |f(x_i)-g(x_i)|-|f(x_{i-1})-g(x_{i-1})| |\leq |[f(x_i)-g(x_i)]-[f(x_{i-1})-g(x_{i-1})]|$$ $$ \leq |f(x_i)-f(x_{i-1})|+|g(x_i)-g(x_{i-1})|$$.