\(Q1\) 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

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

\(Q1\) 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

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

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## Comments

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TopNewestApplying 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 · 1 year ago

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– Satyajit Ghosh · 1 year ago

Thanks. How could I forget to tag you! Btw can you even have a look at Olympiad corner#1 q1Log in to reply

– Shivam Jadhav · 1 year ago

AnsweredLog in to reply

– Satyajit Ghosh · 1 year ago

What about 2nd question?Log in to reply

– Kushagra Sahni · 1 year ago

He is asking for question 1 solution. 2nd was easy.Log in to reply

@Dev Sharma @Adarsh Kumar @Surya Prakash @Svatejas Shivakumar @Kushagra Sahni – Satyajit Ghosh · 1 year ago

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@Satyajit Ghosh Thank you for mentioning me! – Svatejas Shivakumar · 1 year ago

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Therefore, the maximum value of \(y\) is \(\sqrt {8}\) – Svatejas Shivakumar · 1 year ago

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– Satyajit Ghosh · 1 year ago

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

Do u think the 2nd question is correct? Check the expression again from your book. – Kushagra Sahni · 1 year ago

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– Satyajit Ghosh · 1 year ago

If you have got the answer, please give a hint at leastLog in to reply

– Satyajit Ghosh · 1 year ago

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

Lol both questions are Q1? – Kushagra Sahni · 1 year ago

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– Abdur Rehman Zahid · 1 year ago

I've fixed it.Log in to reply