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Q1Q1Q1 Find the largest y such that
11+x2≥y−xy+xforallx>0\frac { 1 }{ 1+{ x }^{ 2 } } \ge \frac { y-x }{ y+x } \quad for\quad all\quad x>01+x21≥y+xy−xforallx>0
Q2Q2Q2 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 } xy+x+1x+1+yz+y+1y+1+zx+z+1z+1
Note by Satyajit Ghosh 5 years, 2 months ago
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@Dev Sharma @Adarsh Kumar @Surya Prakash @Svatejas Shivakumar @Kushagra Sahni
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@Satyajit Ghosh Thank you for mentioning me!
Applying componendo and dividendo we get 2+x2−x2≥2y−2x\frac{2+x^{2}}{-x^{2}}\geq\frac{2y}{-2x}−x22+x2≥−2x2y
2x+x≥y\frac{2}{x}+x \geq y x2+x≥y
Applying A.M−G.MA.M-G.MA.M−G.M
2x+x≥22\frac{2}{x}+x\geq 2\sqrt{2}x2+x≥22
Therefore y=22y=2\sqrt{2}y=22.
Thanks. How could I forget to tag you! Btw can you even have a look at Olympiad corner#1 q1
Answered
@Shivam Jadhav – He is asking for question 1 solution. 2nd was easy.
@Shivam Jadhav – What about 2nd question?
Lol both questions are Q1?
I've fixed it.
Do u think the 2nd question is correct? Check the expression again from your book.
Sorry the denominator of 1st term had xy+y+1 but now I have updated it it to the correct question
If you have got the answer, please give a hint at least
Therefore, the maximum value of yyy is 8\sqrt {8}8
Thanks! Can you tell the answer for q2. Do check out my Olympiad corner#1 which has q1 unanswered.
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Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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or_italics_
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or__bold__
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or\[
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
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Top Newest@Dev Sharma @Adarsh Kumar @Surya Prakash @Svatejas Shivakumar @Kushagra Sahni
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@Satyajit Ghosh Thank you for mentioning me!
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Applying componendo and dividendo we get −x22+x2≥−2x2y
x2+x≥y
Applying A.M−G.M
x2+x≥22
Therefore y=22.
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Thanks. How could I forget to tag you! Btw can you even have a look at Olympiad corner#1 q1
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Answered
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Lol both questions are Q1?
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I've fixed it.
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Do u think the 2nd question is correct? Check the expression again from your book.
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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
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If you have got the answer, please give a hint at least
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Therefore, the maximum value of y is 8
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Thanks! Can you tell the answer for q2. Do check out my Olympiad corner#1 which has q1 unanswered.
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