Olympiad corner #2

Q1Q1 Find the largest y such that

11+x2yxy+xforallx>0\frac { 1 }{ 1+{ x }^{ 2 } } \ge \frac { y-x }{ y+x } \quad for\quad all\quad x>0

Q2Q2 Find the minimum and maximum values of

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

Note by Satyajit Ghosh
4 years ago

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  1. Cross multiplying we get y+xyx1+x2\dfrac{y+x}{y-x} \ge 1+x^{2} or 2yxx\dfrac{2}{y-x} \ge x or x2yx+20x^{2}-yx+2 \le 0.
This is a quadratic equation in xx and since it is 0\le 0, the discriminant i.e. y28y^{2}-8 must also be 0\le 0 or y8y \le \sqrt {8}.

Therefore, the maximum value of yy is 8\sqrt {8}

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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 - 4 years ago

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Applying componendo and dividendo we get 2+x2x22y2x\frac{2+x^{2}}{-x^{2}}\geq\frac{2y}{-2x}

2x+xy\frac{2}{x}+x \geq y

Applying A.MG.MA.M-G.M

2x+x22\frac{2}{x}+x\geq 2\sqrt{2}

Therefore y=22y=2\sqrt{2}.

Shivam Jadhav - 4 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 - 4 years ago

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Answered

Shivam Jadhav - 4 years ago

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@Shivam Jadhav What about 2nd question?

Satyajit Ghosh - 4 years ago

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

Kushagra Sahni - 4 years ago

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

Kushagra Sahni - 4 years ago

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

Satyajit Ghosh - 4 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 - 4 years ago

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Lol both questions are Q1?

Kushagra Sahni - 4 years ago

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I've fixed it.

Abdur Rehman Zahid - 4 years ago

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@Satyajit Ghosh Thank you for mentioning me!

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