Approximating minimum of $f(x)+g(x)$ by solving for $f(x)=g(x)$

If you restrict your attention to nonnegative monotone functions and take $f,g:\mathbb R\to\mathbb R_{\geq0}$ so that $f$ is nondecreasing and $g$ is nonincreasing, suppose we find some $y$ so that $f(y)=g(y)$. Set $M := f(y)+g(y)$ as our "approximate" minimum for the sum.

By the nonnegativity assumption, we will have for $x>y$ that $f(x)+g(x)\geq f(x)\geq f(y) = \frac M2$ and likewise when $x<y$. I'm not sure what the context of this problem is, but this shows that using a solution $f(y)=g(y)$ in this context will always provide a $2$-approximation for the minimum of $f+g$. Since they're only speaking about an "approximate" minimum, maybe this is sufficient for their purposes?

In fact, a $2$-approximation is the best you can do this way: just take $f(x)=c$ and $g(x)$ so that $g(0)=c$ and $g(x)=0$ for $x\neq0$, then $\min(f+g)=c$ but $f(0)+g(0)=2c$.

Moreover, like you mentioned, dropping the nonnegativity constraint makes this even worse. For a concrete example, you can just take $f=0$ and $g(x)=-x$, then this approximation will output $0$ while the true minimum is unbounded below.