If $M(r)$ for $a \leq r \leq b$ by $M(r)=\max\{\frac {r}{a}-1,1-\frac {r}{b}\}$.Then $\min \{M(r):a \leq r \leq b\}=$? I faced the following problem that says:

Let $0<a<b$. Define a function $M(r)$  for $a \leq r \leq b$ by $M(r)=\max\{\frac {r}{a}-1,1-\frac {r}{b}\}$.Then $\min \{M(r):a \leq r \leq b\}$ is which of the following:
$1.0$
$ 2.\frac {2ab}{a+b}$
$3.\frac {b-a}{b+a}$
$4.\frac {b+a}{b-a}$

I do not know how to progress with it.Can someone point me in the right direction? Thanks in advance for your time.

If $a\leq r \leq \dfrac{2ab}{a+b}$ show that $M(r)=1-\dfrac{r}{b}$ and for $\dfrac{2ab}{a+b}\leq r \leq b$ that $M(r)=\dfrac{r}{a}-1$. 
As Did suggested a picture (of $\dfrac{r}{a}-1, 1-\dfrac{r}{b}$) will help. What is $\dfrac{2ab}{a+b}$?
To find $\min \{M(r):a\leq r\leq b\}$ note that $\dfrac{r}{a}-1$ is increasing and $1-\dfrac{r}{b}$ decreasing.
