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, 12 months ago

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– Nihar Mahajan · 1 year, 12 months ago

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

– Alan Yan · 1 year, 12 months ago

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

– Nihar Mahajan · 1 year, 12 months ago

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

– Alan Yan · 1 year, 12 months ago

:) Nice.Log in to reply

Click here :) – Nihar Mahajan · 1 year, 12 months 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, 12 months ago

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

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– Nihar Mahajan · 1 year, 12 months 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, 12 months 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, 12 months ago

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– Saarthak Marathe · 1 year, 12 months 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, 11 months ago

Titu is such a cute name lolLog in to reply

– Alan Yan · 1 year, 11 months 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, 12 months 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, 12 months 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, 12 months ago

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