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Prove that cyclic sum of $\displaystyle \sum_{\text{cyclic}} \dfrac{a^3}{a^2+ab+b^2} \geq \dfrac{a+b+c}{3}$ , if $a, b, c > 0$

I'm really stuck on this one. Tried some stuff involving QM> AM(because the are positive) but can't derive the needed ,can't proceed from it.

DeepSea
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    It'd help to explain what the cyclic sum is, because that's a notation that's only used by people preparing for math contests and isn't familiar to mainstream mathematicians. – Daniel McLaury Oct 19 '14 at 05:32
  • I think it should be $a^3$ not $a^2$ in the denominator. – Theo Oct 19 '14 at 05:38
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    Bring to the same common denominator, expand, and this should follow from Muirhead Inequality. – Theo Oct 19 '14 at 05:48

4 Answers4

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Note that $$\frac{a^3-b^3}{a^2+ab+b^2}=a-b \implies LHS = \sum_{cyc} \frac{a^3}{a^2+ab+b^2} = \sum_{cyc} \frac{b^3}{a^2+ab+b^2}$$

So we get $$2LHS = \sum_{cyc} \frac{a^3+b^3}{a^2+ab+b^2}$$

But by rearrangement, $a^3+b^3\ge a^2b+ab^2 \implies 3a^3+3b^3 \ge a^3+2a^2b+2ab^2+b^3 = (a+b)(a^2+ab+b^2)$. $$\therefore 2LHS = \sum_{cyc} \frac{a^3+b^3}{a^2+ab+b^2} \ge \frac13\sum_{cyc} (a+b)= \frac23(a+b+c)$$ which is what we wanted to show.

Macavity
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Here is another way, using Cauchy-Schwarz inequality: $$\sum_{cyc} \frac{a^4}{a(a^2+ab+b^2)} \ge \frac{(a^2+b^2+c^2)^2}{\sum_{cyc} a(a^2+ab+b^2)} = \frac{(a^2+b^2+c^2)^2}{(a+b+c)(a^2+b^2+c^2)}$$

So it is enough to show that $$a^2+b^2+c^2 \ge \frac{(a+b+c)^2}3$$ which is easy again using Cauchy-Schwarz.

Macavity
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First, we find :

$$\sum {{{{a^3}} \over {{a^2} + ab + {b^2}}}} = \sum {{{{a^3} - {b^3} + {b^3}} \over {{a^2} + ab + {b^2}}}} = \sum {\left( {a - b} \right)} + \sum {{{{b^3}} \over {{a^2} + ab + {b^2}}}} = \sum {{{{b^3}} \over {{a^2} + ab + {b^2}}}} $$

So :

$$\sum {{{{a^3}} \over {{a^2} + ab + {b^2}}}} = {1 \over 2}\sum {{{{a^3} + {b^3}} \over {{a^2} + ab + {b^2}}}} $$

Since : $${\left( {a - b} \right)^2} \ge 0 \Longrightarrow 2{a^2} + 2{b^2} - 4ab \ge 0 \Longrightarrow 3\left( {{a^2} - ab + {b^2}} \right) \ge {a^2} + ab + {b^2} \\ \Longrightarrow {{{a^2} - ab + {b^2}} \over {{a^2} + ab + {b^2}}} \ge {1 \over 3} \Longrightarrow {1 \over 2} \cdot {{{a^3} + {b^3}} \over {{a^2} + ab + {b^2}}} \ge {{a + b} \over 6} $$

So we get: $$ \sum {{{{a^3}} \over {{a^2} + ab + {b^2}}}} = {1 \over 2}\sum {{{{a^3} + {b^3}} \over {{a^2} + ab + {b^2}}}} \ge {{a + b + c} \over 3}$$

Shine Mic
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We need to prove that $$\sum_{cyc}\left(\frac{a^3}{a^2+ab+b^2}-\frac{a}{3}\right)\geq0$$ or $$\sum_{cyc}\frac{a(2a+b)(a-b)}{a^2+ab+b^2}\geq0$$ or $$\sum_{cyc}\left(\frac{a(2a+b)(a-b)}{a^2+ab+b^2}-(a-b)\right)\geq0$$ or $$\sum_{cyc}\frac{(a-b)^2(a+b)}{a^2+ab+b^2}\geq0.$$ Done!

Michael Rozenberg
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