RMO Inequality practice (2) (Original)

If a,b,ca,b,c are positive reals and nn is a positive integer , Prove the following inequality:

nab+nc+nbc+na+nca+nb3nn+1\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
4 years, 1 month ago

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Prove a3a2+ab+b2+b3b2+bc+c2+c3c2+ac+a23abcab+bc+ca{\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 - 4 years, 1 month ago

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

Nihar Mahajan - 4 years, 1 month ago

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

Alan Yan - 4 years, 1 month 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 - 4 years, 1 month ago

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@Nihar Mahajan :) Nice.

Alan Yan - 4 years, 1 month ago

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@Alan Yan Click here :)

Nihar Mahajan - 4 years, 1 month ago

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cyca3a2+ab+b2=cyca4a(a2+ab+b2)(cyca2)cyca(a2+ab+b2)=(a2+b2+c2)2(a+b+c)(a2+b2+c2)=a2+b2+c2a+b+c(a+b+c)23(a+b+c)=a+b+c3\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:

a+b+c33abcab+bc+ac(a+b+c)(ab+bc+ac)9abc\large{\dfrac{a+b+c}{3} \geq \dfrac{3abc}{ab+bc+ac} \Rightarrow (a+b+c)(ab+bc+ac) \geq 9abc}

Proof:

(a+b+c)(ab+bc+ac)=cyc(a2b+abc+a2c)=3abc+cyca2(b+c)=3abc+abc(cyca(b+c)bc)=3abc+abc(cyc(ab+ba))3abc+6abc=9abc\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 - 4 years, 1 month ago

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By T2's Lemma cycnab+ncn2(a+b+c)2(ab+bc+ca)(n2+n)=n(a+b+c)2(n+1)(ab+bc+ca)3nn+1\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 - 4 years, 1 month ago

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

Given that a2+b2+c2=1a^2+b^2+c^2=1, prove that nab+nc+nbc+na+nca+nb3nn+3(a3b+b3c+c3a)\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 - 4 years, 1 month ago

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

Nihar Mahajan - 4 years, 1 month 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".

Saarthak Marathe - 4 years, 1 month ago

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Titu is such a cute name lol

Nihar Mahajan - 4 years, 1 month 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 - 4 years, 1 month ago

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

Daniel Liu - 4 years, 1 month ago

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Wrong for n=1 n = -1 :P

Alan Yan - 4 years, 1 month ago

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Thanks edited.

Nihar Mahajan - 4 years, 1 month ago

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

Nihar Mahajan - 4 years, 1 month ago

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