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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".

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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

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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:

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

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

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

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Click here :)

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Wrong for $n = -1$ :P

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Thanks edited.

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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.

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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".

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Titu is such a cute name lol

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

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

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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 $\dfrac{na}{b+nc}+\dfrac{nb}{c+na}+\dfrac{nc}{a+nb}\ge \dfrac{3n}{n+3(a^3b+b^3c+c^3a)}$

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

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