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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
10 months, 2 weeks 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 · 10 months, 2 weeks ago

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@Shivam Jadhav Thanks. How could I forget to tag you! Btw can you even have a look at Olympiad corner#1 q1 Satyajit Ghosh · 10 months, 2 weeks ago

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@Satyajit Ghosh Answered Shivam Jadhav · 10 months, 2 weeks ago

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@Shivam Jadhav What about 2nd question? Satyajit Ghosh · 10 months, 2 weeks ago

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@Shivam Jadhav He is asking for question 1 solution. 2nd was easy. Kushagra Sahni · 10 months, 2 weeks ago

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@Satyajit Ghosh @Satyajit Ghosh Thank you for mentioning me! Svatejas Shivakumar · 10 months, 2 weeks 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 \(\sqrt {8}\) Svatejas Shivakumar · 10 months, 2 weeks ago

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@Svatejas Shivakumar Thanks! Can you tell the answer for q2. Do check out my Olympiad corner#1 which has q1 unanswered. Satyajit Ghosh · 10 months, 2 weeks ago

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Do u think the 2nd question is correct? Check the expression again from your book. Kushagra Sahni · 10 months, 2 weeks ago

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@Kushagra Sahni If you have got the answer, please give a hint at least Satyajit Ghosh · 10 months, 2 weeks ago

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@Kushagra Sahni Sorry the denominator of 1st term had xy+y+1 but now I have updated it it to the correct question Satyajit Ghosh · 10 months, 2 weeks ago

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Lol both questions are Q1? Kushagra Sahni · 10 months, 2 weeks ago

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@Kushagra Sahni I've fixed it. Abdur Rehman Zahid · 10 months, 2 weeks ago

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