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RMO Inequality practice (2) (Original)

If \(a,b,c\) are positive reals and \(n\) is a positive integer , Prove the following inequality:

\[\large\dfrac{na}{b+nc} + \dfrac{nb}{c+na} +\dfrac{nc}{a+nb} \geq \dfrac{3n}{n+1}\]


This problem is original and is inspired by some note.

Note by Nihar Mahajan
1 year, 12 months ago

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Prove \[{\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+a^2}}\geq{\frac{3abc}{ab+bc+ca}} \] Alan Yan · 1 year, 12 months ago

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@Alan Yan \(a,b,c\) are positive reals right? Nihar Mahajan · 1 year, 12 months ago

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@Nihar Mahajan Yes. It is a nice problem :) But there is a stronger one after. Alan Yan · 1 year, 12 months ago

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@Alan Yan I got inspired by this problem. I am posting something (its easy though). Stay tuned :P Nihar Mahajan · 1 year, 12 months ago

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@Nihar Mahajan :) Nice. Alan Yan · 1 year, 12 months ago

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@Alan Yan Click here :) Nihar Mahajan · 1 year, 12 months ago

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@Alan Yan \[\large {\sum_{cyc} \dfrac{a^3}{a^2+ab+b^2} \\ = \sum_{cyc} \dfrac{a^4}{a(a^2+ab+b^2)} \\ \geq \dfrac{\left(\displaystyle\sum_{cyc} a^2\right)}{\displaystyle\sum_{cyc} a(a^2+ab+b^2)} \\ = \dfrac{(a^2+b^2+c^2)^2}{(a+b+c)(a^2+b^2+c^2)} \\ = \dfrac{a^2+b^2+c^2}{a+b+c} \\ \geq \dfrac{(a+b+c)^2}{3(a+b+c)} \\ = \dfrac{a+b+c}{3} }\]

Thus it suffices to prove that:

\(\large{\dfrac{a+b+c}{3} \geq \dfrac{3abc}{ab+bc+ac} \Rightarrow (a+b+c)(ab+bc+ac) \geq 9abc}\)

Proof:

\[\large{(a+b+c)(ab+bc+ac) \\ = \sum_{cyc} (a^2b+abc+a^2c) \\ = 3abc + \sum_{cyc} a^2(b+c) \\ = 3abc+abc\left(\sum_{cyc} \dfrac{a(b+c)}{bc}\right) \\ = 3abc+abc\left(\sum_{cyc} \left(\dfrac{a}{b}+\dfrac{b}{a}\right)\right) \\ \geq 3abc+6abc = 9abc} \] Nihar Mahajan · 1 year, 12 months ago

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Wrong for \( n = -1\) :P Alan Yan · 1 year, 12 months ago

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@Alan Yan Thanks edited. Nihar Mahajan · 1 year, 12 months ago

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By T2's Lemma \[\sum_{cyc}\dfrac{na}{b+nc}\ge \dfrac{n^2(a+b+c)^2}{(ab+bc+ca)(n^2+n)}=\dfrac{n(a+b+c)^2}{(n+1)(ab+bc+ca)}\ge \dfrac{3n}{n+1}\] done. Daniel Liu · 1 year, 12 months ago

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@Daniel Liu Here's a strengthening (albeit not a very good one):

Given that \(a^2+b^2+c^2=1\), prove that \[\dfrac{na}{b+nc}+\dfrac{nb}{c+na}+\dfrac{nc}{a+nb}\ge \dfrac{3n}{n+3(a^3b+b^3c+c^3a)}\] Daniel Liu · 1 year, 12 months ago

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@Daniel Liu Done the same way. And by the way,the inequality was not first found by Titu (neither by Arthur Engel),it was found by some other Russian mathematician. It is mentioned in "Kvant". Saarthak Marathe · 1 year, 12 months ago

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@Saarthak Marathe Titu is such a cute name lol Nihar Mahajan · 1 year, 11 months ago

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@Nihar Mahajan Titu Andresscu actually has A LOT of good books about every topic of math in olympiads. :) Alan Yan · 1 year, 11 months ago

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@Saarthak Marathe That is true, but still most people call it T2's lemma or Engel form of CS. Daniel Liu · 1 year, 12 months ago

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@Daniel Liu Yay! I got inspired by this note quite lucid it is. Nihar Mahajan · 1 year, 12 months ago

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@Calvin Lin @Harsh Shrivastava @Alan Yan @Saarthak Marathe Hope all enjoy solving it :) Nihar Mahajan · 1 year, 12 months ago

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