1. ## Analysis inequality

Let a, b be two positive real numbers. Show that arctan((a+1)/b) - arctan(a/b) < b/(a^2 + b^2)

2. This looks like a mean value theorem question.

It's too late for me to try it now, i'll have a go in the morning.

3. Originally Posted by kjwill1776
Let a, b be two positive real numbers. Show that $\displaystyle \arctan\left(\frac{a+1}{b}\right)-\arctan\left(\frac{a}{b}\right)<\frac{b}{a^2+b^2}$
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:
Thus, $\displaystyle \frac{\arctan\left(\frac{a}{b}+\frac{1}{b}\right)-\arctan\left(\frac{a}{b}\right)}{\frac{1}{b}}<\fra c{1}{1+\left(\frac{a}{b}\right)^2} = \frac{b^2}{a^2+b^2}$

Multiplying both sides by $\displaystyle \frac{1}{b}$ gives us:

$\displaystyle \arctan\left(\frac{a}{b}+\frac{1}{b}\right)-\arctan\left(\frac{a}{b}\right)<\frac{b}{a^2+b^2}$