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Thread: Analysis inequality

  1. #1
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    Analysis inequality

    Let a, b be two positive real numbers. Show that arctan((a+1)/b) - arctan(a/b) < b/(a^2 + b^2)
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  2. #2
    Super Member Showcase_22's Avatar
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    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.
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by kjwill1776 View Post
    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:

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