\(Q1\) 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
\(\frac { x+1 }{ xy+x+1 } +\frac { y+1 }{ yz+y+1 } +\frac { z+1 }{ zx+z+1 } \)
Note by
Satyajit Ghosh
1 year, 8 months ago
Problem Loading...
Note Loading...
Set Loading...
Comments
Sort by:
Top NewestApplying 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, 8 months ago
Log in to reply
Thanks. How could I forget to tag you! Btw can you even have a look at Olympiad corner#1 q1 – Satyajit Ghosh · 1 year, 8 months ago
Log in to reply
Answered – Shivam Jadhav · 1 year, 8 months ago
Log in to reply
What about 2nd question? – Satyajit Ghosh · 1 year, 8 months ago
Log in to reply
He is asking for question 1 solution. 2nd was easy. – Kushagra Sahni · 1 year, 8 months ago
Log in to reply
@Dev Sharma @Adarsh Kumar @Surya Prakash @Svatejas Shivakumar @Kushagra Sahni – Satyajit Ghosh · 1 year, 8 months ago
Log in to reply
@Satyajit Ghosh Thank you for mentioning me! – Svatejas Shivakumar · 1 year, 8 months ago
Log in to reply
- 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 · 1 year, 8 months ago
Log in to reply
Thanks! Can you tell the answer for q2. Do check out my Olympiad corner#1 which has q1 unanswered. – Satyajit Ghosh · 1 year, 8 months ago
Log in to reply
Do u think the 2nd question is correct? Check the expression again from your book. – Kushagra Sahni · 1 year, 8 months ago
Log in to reply
If you have got the answer, please give a hint at least – Satyajit Ghosh · 1 year, 8 months ago
Log in to reply
Sorry the denominator of 1st term had xy+y+1 but now I have updated it it to the correct question – Satyajit Ghosh · 1 year, 8 months ago
Log in to reply
Lol both questions are Q1? – Kushagra Sahni · 1 year, 8 months ago
Log in to reply
I've fixed it. – Abdur Rehman Zahid · 1 year, 8 months ago
Log in to reply