$Q1$ 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
4 years, 11 months ago

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- 4 years, 11 months ago

@Satyajit Ghosh Thank you for mentioning me!

- 4 years, 11 months ago

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 $y=2\sqrt{2}$.

- 4 years, 11 months ago

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

- 4 years, 11 months ago

- 4 years, 11 months ago

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

- 4 years, 11 months ago

- 4 years, 11 months ago

Lol both questions are Q1?

- 4 years, 11 months ago

I've fixed it.

- 4 years, 11 months ago

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

- 4 years, 11 months ago

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

- 4 years, 11 months ago

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

- 4 years, 11 months ago

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 $\sqrt {8}$

- 4 years, 11 months ago

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

- 4 years, 11 months ago