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$$Q1$$ 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, 4 months 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 $$y=2\sqrt{2}$$. · 1 year, 4 months ago

Thanks. How could I forget to tag you! Btw can you even have a look at Olympiad corner#1 q1 · 1 year, 4 months ago

Answered · 1 year, 4 months ago

What about 2nd question? · 1 year, 4 months ago

He is asking for question 1 solution. 2nd was easy. · 1 year, 4 months ago

@Satyajit Ghosh Thank you for mentioning me! · 1 year, 4 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 $$\sqrt {8}$$ · 1 year, 4 months ago

Thanks! Can you tell the answer for q2. Do check out my Olympiad corner#1 which has q1 unanswered. · 1 year, 4 months ago

Do u think the 2nd question is correct? Check the expression again from your book. · 1 year, 4 months ago

If you have got the answer, please give a hint at least · 1 year, 4 months ago

Sorry the denominator of 1st term had xy+y+1 but now I have updated it it to the correct question · 1 year, 4 months ago

Lol both questions are Q1? · 1 year, 4 months ago