Absolute value of difference of minimums of two functions is bounded above by their maximum absolute difference

Question

If $f$ and $g$ are two real valued positive functions. I need to prove the inequality $\max\limits_x|f(x)-g(x)|\geq |\min\limits_x f(x) - \min\limits_x g(x)| $. I am not able to get any counter example disproving the statement.

Answer 1

WLOG, assume that $f(x_1)=\min f(x)>\min g(x)=g(x_2)$, then $$\max |f(x)-g(x)|\geq |f(x_2)-g(x_2)|\geq f(x_2)-g(x_2)\geq f(x_1)-g(x_2)$$

Answer 2

Without loss of generality suppose $\min_x f(x) \ge \min_x g(x)$. Let $x_0$ be such that $g(x_0) = \min_x g(x)$. Then $$|\min_x f(x) - \min_x g(x)| = \min_x f(x) - g(x_0) \le f(x_0) - g(x_0) \le \max_x |f(x) - g(x)|.$$

(I was a little un-rigorous in assuming that $g$ attains its infimum, but the argument can be modified appropriately.)