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

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

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

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

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

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– Nihar Mahajan · 1 year ago

\(a,b,c\) are positive reals right?Log in to reply

– Alan Yan · 1 year ago

Yes. It is a nice problem :) But there is a stronger one after.Log in to reply

– Nihar Mahajan · 1 year ago

I got inspired by this problem. I am posting something (its easy though). Stay tuned :PLog in to reply

– Alan Yan · 1 year ago

:) Nice.Log in to reply

Click here :) – Nihar Mahajan · 1 year ago

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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 ago

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

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– Nihar Mahajan · 1 year ago

Thanks edited.Log in to reply

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 ago

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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 ago

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– Saarthak Marathe · 1 year ago

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".Log in to reply

– Nihar Mahajan · 1 year ago

Titu is such a cute name lolLog in to reply

– Alan Yan · 1 year ago

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

– Daniel Liu · 1 year ago

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

this note quite lucid it is. – Nihar Mahajan · 1 year ago

Yay! I got inspired byLog in to reply

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

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