Show that a simple inequality holds: $a<b$ implies $\frac ab < \frac{a+1}{b+1}$

I'm trying to show that $\dfrac{a}{b} < \dfrac{a + 1}{b + 1}$ given $a < b$ and $b$ is positive.

Ideas?

Consider the difference

$$\frac{a + 1}{b + 1} - \frac{a}{b} = \frac{(a + 1)b - a(b + 1)}{b(b + 1)} = \frac{b - a}{b(b + 1)} > 0$$

Here is a geometrical explanation in terms of slopes.

If take $b$ for abscissa and $a$ as ordinate of a certain point $M$, adding vector $\binom{1}{1}$ to vector $\vec{OM}=\binom{b}{a}$ will an increase of the slope (due to the fact that $a>b$) as illustrated on the following figure: Fig. 1: $(a+1)/(b+1) > a/b$ is equivalent to the fact that slope $ON$ > slope $OM$.

Remark: moreover the signed area of triangle $OMN$ is the half of $\det(\vec{OM},\vec{ON})$ (in correspondence with $\vec{OM} \times \vec{ON}$.