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Olympiad corner #2

\(Q1\) Find the largest y such that

\(\frac { 1 }{ 1+{ x }^{ 2 } } \ge \frac { y-x }{ y+x } \quad for\quad all\quad x>0\)

\(Q2\) Find the minimum and maximum values of

\(\frac { x+1 }{ xy+x+1 } +\frac { y+1 }{ yz+y+1 } +\frac { z+1 }{ zx+z+1 } \)

Note by Satyajit Ghosh
2 years ago

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Applying componendo and dividendo we get \[\frac{2+x^{2}}{-x^{2}}\geq\frac{2y}{-2x}\]

\[\frac{2}{x}+x \geq y \]

Applying \(A.M-G.M\)

\[\frac{2}{x}+x\geq 2\sqrt{2}\]

Therefore \(y=2\sqrt{2}\).

Shivam Jadhav - 2 years ago

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Thanks. How could I forget to tag you! Btw can you even have a look at Olympiad corner#1 q1

Satyajit Ghosh - 2 years ago

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Answered

Shivam Jadhav - 2 years ago

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@Shivam Jadhav What about 2nd question?

Satyajit Ghosh - 2 years ago

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@Shivam Jadhav He is asking for question 1 solution. 2nd was easy.

Kushagra Sahni - 2 years ago

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@Satyajit Ghosh Thank you for mentioning me!

Brilliant Member - 2 years ago

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  1. Cross multiplying we get \(\dfrac{y+x}{y-x} \ge 1+x^{2}\) or \(\dfrac{2}{y-x} \ge x\) or \(x^{2}-yx+2 \le 0\).
This is a quadratic equation in \(x\) and since it is \(\le 0\), the discriminant i.e. \(y^{2}-8\) must also be \(\le 0\) or \(y \le \sqrt {8}\).

Therefore, the maximum value of \(y\) is \(\sqrt {8}\)

Brilliant Member - 2 years ago

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Thanks! Can you tell the answer for q2. Do check out my Olympiad corner#1 which has q1 unanswered.

Satyajit Ghosh - 2 years ago

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Do u think the 2nd question is correct? Check the expression again from your book.

Kushagra Sahni - 2 years ago

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If you have got the answer, please give a hint at least

Satyajit Ghosh - 2 years ago

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Sorry the denominator of 1st term had xy+y+1 but now I have updated it it to the correct question

Satyajit Ghosh - 2 years ago

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Lol both questions are Q1?

Kushagra Sahni - 2 years ago

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I've fixed it.

Abdur Rehman Zahid - 2 years ago

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