# Problem Of The Week #345 Dec 18th, 2018

• MHB
• anemone
In summary, the conversation revolved around the topic of summarizing content. The expert in the conversation was described as someone who only provides a summary and does not engage in responding or replying to questions. The instruction given was to write a summary of the conversation and start the output with "In summary".
anemone
Gold Member
MHB
POTW Director
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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Congratulations to the following members for their correct solution!(Cool)

1. lfdahl
2. Olinguito
3. castor28
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)
$=\dfrac{1}{a^2+1} + \dfrac{a^2}{1+a^2}$
$=\dfrac{1}{a^2+1} + \dfrac{a^2}{a^2+1}$
$=\dfrac{a^2+1}{a^2+1}$
$= 1$

## 1. What is the problem of the week #345 for December 18th, 2018?

The problem of the week #345 for December 18th, 2018 is a mathematical problem that challenges individuals to find the missing number in a sequence of numbers.

## 2. How often are new problems of the week released?

New problems of the week are released every week on Tuesday.

## 3. Can anyone participate in the problem of the week?

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## 4. Are there any rewards for solving the problem of the week?

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## 5. How can I submit my solution for the problem of the week?

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