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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
2 years, 5 months ago

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## Comments

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Prove ${\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+a^2}}\geq{\frac{3abc}{ab+bc+ca}}$

- 2 years, 5 months ago

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$$a,b,c$$ are positive reals right?

- 2 years, 5 months ago

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Yes. It is a nice problem :) But there is a stronger one after.

- 2 years, 5 months ago

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I got inspired by this problem. I am posting something (its easy though). Stay tuned :P

- 2 years, 5 months ago

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:) Nice.

- 2 years, 5 months ago

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Click here :)

- 2 years, 5 months ago

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$\large {\sum_{cyc} \dfrac{a^3}{a^2+ab+b^2} \\ = \sum_{cyc} \dfrac{a^4}{a(a^2+ab+b^2)} \\ \geq \dfrac{\left(\displaystyle\sum_{cyc} a^2\right)}{\displaystyle\sum_{cyc} a(a^2+ab+b^2)} \\ = \dfrac{(a^2+b^2+c^2)^2}{(a+b+c)(a^2+b^2+c^2)} \\ = \dfrac{a^2+b^2+c^2}{a+b+c} \\ \geq \dfrac{(a+b+c)^2}{3(a+b+c)} \\ = \dfrac{a+b+c}{3} }$

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:

$\large{(a+b+c)(ab+bc+ac) \\ = \sum_{cyc} (a^2b+abc+a^2c) \\ = 3abc + \sum_{cyc} a^2(b+c) \\ = 3abc+abc\left(\sum_{cyc} \dfrac{a(b+c)}{bc}\right) \\ = 3abc+abc\left(\sum_{cyc} \left(\dfrac{a}{b}+\dfrac{b}{a}\right)\right) \\ \geq 3abc+6abc = 9abc}$

- 2 years, 5 months ago

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Wrong for $$n = -1$$ :P

- 2 years, 5 months ago

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Thanks edited.

- 2 years, 5 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.

- 2 years, 5 months ago

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

Given that $$a^2+b^2+c^2=1$$, prove that $\dfrac{na}{b+nc}+\dfrac{nb}{c+na}+\dfrac{nc}{a+nb}\ge \dfrac{3n}{n+3(a^3b+b^3c+c^3a)}$

- 2 years, 5 months ago

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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".

- 2 years, 5 months ago

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Titu is such a cute name lol

- 2 years, 5 months ago

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Titu Andresscu actually has A LOT of good books about every topic of math in olympiads. :)

- 2 years, 5 months ago

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That is true, but still most people call it T2's lemma or Engel form of CS.

- 2 years, 5 months ago

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Yay! I got inspired by this note quite lucid it is.

- 2 years, 5 months ago

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@Calvin Lin @Harsh Shrivastava @Alan Yan @Saarthak Marathe Hope all enjoy solving it :)

- 2 years, 5 months ago

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