$\sup_{x \in [a,b]} f - \inf_{x \in [a,b]} f = \sup_{x,y \in [a,b]} |f(x) - f(y)|$ It seems somehow intuitive. But how could it be done?

Prove that if $f$ is a bounded function on $[a,b] \subset \mathbb R$, then:
$\sup_{x \in [a,b]} f - \inf_{x \in [a,b]} f = \sup_{x,y \in [a,b]} |f(x) - f(y)|$

Thank you.
 A: Let 
$$\lambda = \sup \{ \ f(x) \ \mid \ x \in [a, b]\}$$
$$\mu = \inf \{ \ f(x) \ \mid \ x \in [a, b]\}$$
$$\eta = \sup \{ \ f(x) - f(y) \ \mid \ x, y \in [a, b]\} $$
All the suprema and infima above are defined since $f$ is bounded on $[a, b]$ but $f$ does not necessarily have to attain those values. The equality still holds. And, notice I did not consider the absolute value of $f(x) - f(y)$ but the two suprema are equal and this one is easier to work with. 
First notice that for each $x, y \in [a, b]$
$$ f(x) - f(y) \le \lambda - f(y) \le \lambda - \mu$$
Hence, $$\eta \le \lambda - \mu \tag{1}$$
And suppose for teh sake of a contradiction that $\eta \lt \lambda - \mu$. Then $\eta + \mu \lt \lambda$. Then since $\lambda  = \sup {f(x)}$ there exists $x\in [a, b]$ such that $\eta + \mu \lt f(x)$. Then $\mu \lt f(x) - \eta$. Now since $\mu = \inf{f(x)}$ there must be a $y \in [a, b]$ such that $f(y) \lt f(x) - \eta$. But then $\eta \lt f(x) - f(y)$ which contradicts the definition of $\eta$. 
Hence from $(1)$ we have that $$\eta = \lambda - \mu  $$
$\mathscr{Q.E.D.}$
A: To get the other inequality, we have that $f(x) \leq f(k_1)$ and $f(y)\geq f(k_2)$ for all $x,y\in [a,b]$. So, 
$$
 f(x)-f(y) \leq f(k_1)-f(k_2).
$$
But we also have that $f(x) \geq f(k_2)$ and $f(y) \leq f(k_1)$. So,
$$
  f(y)-f(x) \leq f(k_1) - f(k_2).
$$
We then have that $|f(x)-f(y)|\leq f(k_1)-f(k_2)$, for arbitrary $x,y\in [a,b]$. The desired inequality follows.
