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Math Help - 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 \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<z, \frac{d}{dx}[\arctan(z)]<\frac{d}{dx}[\arctan(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}<c<\frac{a}{b}+\frac{1}{b}

    From above though, we know that \frac{a}{b}<c 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}
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