Inequality involving floor function and fractions I have little to no experience working with floor inequalities so I am kind of stuck on this one. It seems pretty intuitive though. 
So basically I want to show that
$$\left\lfloor\frac{a}{b}\right\rfloor \geq \left\lfloor\frac{a}{b+1}\right\rfloor$$ where $a$ and $b$ are strictly positive integers.
This is true for $b \geq a$, so I'm looking at the $b < a$ case but have no idea where to start. Primarily the denominators are different and I have no idea how to get it out of the floor function. Thanks!

I think it's obvious because 
$$\frac{a}{b}\gt\frac{a}{b+1}\Rightarrow\left\lfloor\frac ab\right\rfloor\ge\left\lfloor\frac {a}{b+1}\right\rfloor.$$
