Problem Of The Week #345 Dec 18th, 2018

[anemone]

Here is this week's POTW:

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Let $a$ and $b$ are positive real numbers such that $\dfrac{1}{a+1}+\dfrac{1}{b+1}=1.$ Prove that $\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}=1$`

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. lfdahl
2. Olinguito
3. castor28
4. kaliprasadSolution from kaliprasad:

Spoiler

From the given condition multiplying by $(a+1)(b+1)$ we get

$(b+1) + (a+1) = (b+1)(a+1))$

or

b +1 + a + 1 = ab + a + b + 1 or $ab = 1$

or $b= \dfrac{1}{a}\cdots(1)$

Now

$\dfrac{1}{a^2+1} + \dfrac{1}{b^2+1}$
$=\dfrac{1}{a^2+1} + \dfrac{1}{\frac{1}{a^2} + 1}$ from (1)
$=\dfrac{1}{a^2+1} + \dfrac{a^2}{1+a^2}$
$=\dfrac{1}{a^2+1} + \dfrac{a^2}{a^2+1}$
$=\dfrac{a^2+1}{a^2+1} $
$= 1$