# Math Help - Analysis inequality

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 $\arctan\left(\frac{a+1}{b}\right)-\arctan\left(\frac{a}{b}\right)<\frac{b}{a^2+b^2}$
The first thing to note is that $\frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2}$ is a decreasing function. Hence if $y

Now use the Mean Value Theorem:

$\frac{\arctan\left(\frac{a}{b}+\frac{1}{b}\right)-\arctan\left(\frac{a}{b}\right)}{\frac{1}{b}} = \frac{1}{1+c^2}$ for $\frac{a}{b}

From above though, we know that $\frac{a}{b} implies $\frac{1}{1+c^2}<\frac{1}{1+\left(\frac{a}{b}\right )^2}$

The rest is below if you need it.

Spoiler:
Thus, $\frac{\arctan\left(\frac{a}{b}+\frac{1}{b}\right)-\arctan\left(\frac{a}{b}\right)}{\frac{1}{b}}<\fra c{1}{1+\left(\frac{a}{b}\right)^2} = \frac{b^2}{a^2+b^2}$

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

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