How can I prove $(\min_x f(x)) + g(x) \geq \min_x (f(x) + g(x))$?

Question

How can I prove $(\min_x f(x)) + g(x) \geq \min_x (f(x) + g(x))$? In other words, on the left side we only optimize $x$ with respect to function $f(\cdot)$ while on the right side we do so with respect to summation of both functions $f(\cdot)$ and $g(\cdot)$?

Answer 1

This isn't necessarily the case: consider $f(x) = x^2, g(x) = (x - 1)^2.$ Evaluating the left side at $x = 1$ yields $\min[x^2] + (1 - 1)^2 = 0 + 0 = 0,$ but the function $x^2 + (x - 1)^2$ is strictly positive in the real numbers, so $\min [x^2 + (x - 1)^2] > 0.$

The reason this doesn't work out is because on the left-hand side we can effectively pick the values of $x$ for $f$ and $g$ separately, so the minimum value we can achieve is $\min [f(x)] + \min[g(x)],$ but on the left-hand side we must use the same value of $x$ for both $f$ and $g,$ so if the minima don't align then we can't get as low.