Let a, b be two positive real numbers. Show that arctan((a+1)/b) - arctan(a/b) < b/(a^2 + b^2)
The first thing to note is that $\displaystyle \frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2}$ is a decreasing function. Hence if $\displaystyle y<z, \frac{d}{dx}[\arctan(z)]<\frac{d}{dx}[\arctan(y)]$
Now use the Mean Value Theorem:
$\displaystyle \frac{\arctan\left(\frac{a}{b}+\frac{1}{b}\right)-\arctan\left(\frac{a}{b}\right)}{\frac{1}{b}} = \frac{1}{1+c^2}$ for $\displaystyle \frac{a}{b}<c<\frac{a}{b}+\frac{1}{b}$
From above though, we know that $\displaystyle \frac{a}{b}<c$ implies $\displaystyle \frac{1}{1+c^2}<\frac{1}{1+\left(\frac{a}{b}\right )^2}$
The rest is below if you need it.
Spoiler: