It had been long days I had posted Olympiad Proof Problem - Day 5. Sorry for the inconvenience made to those who are waiting for the Proof Problem Day - 6. So, Finally here is the proof problem day - 6.

If \(x_{1}\), \(x_{2}\), \(x_{3}\), \(\ldots\) \(x_{n}\) are positive real numbers, then prove the following inequality

\[\Large \sum_{i=1}^{n} \dfrac{x_{i} ^3}{x_{i}^2 + x_{i}x_{i+1} + x_{i+1}^2} \geq \dfrac{x_{1} + x_{2} + x_{3} + \ldots + x_{n}}{3}\]

**Details and Assumptions:**

- \(x_{n+1} = x_{1}\).

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## Comments

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TopNewestObserve that

\[0 = (x_1 - x_2) + (x_2 - x_3) + \ldots + (x_n - x_1) = \displaystyle \sum_{i=1}^n \dfrac {x_{i}^3 - x_{i+1}^3}{x_i^2+x_i x_{i+1}+x_{i+1}^2} \]

Hence,

\[\displaystyle \sum_{i=1}^n \dfrac {x_i^3}{x_i^2+x_ix_{i+1}+x_{i+1}^2} = \dfrac {1}{2} \displaystyle \sum_{i=1}^n \dfrac {x_i^3+x_{i+1}^3}{x_i^2+x_ix_{i+1}+x_{i+1}^2}\]

Note that \(a^3 + b^3 \geq \frac{1}{3} a^3 + \frac{2}{3} a^2b + \frac {2}{3} ab^2 + \frac{1}{3} b^3 = \frac {1}{3} (a+b)(a^2+ab+b^2)\) (which can be easily proven by AM-GM). From this, we attain:

\[\dfrac {1}{2} \displaystyle \sum_{i=1}^n \dfrac {x_i^3+x_{i+1}^3}{x_i^2+x_ix_{i+1}+x_{i+1}^2} \geq \dfrac {1}{2} \displaystyle \sum_{i=1}^n \dfrac {x_1+x_{i+1}}{3}=\dfrac {1}{3} \displaystyle \sum_{i=1}^n x_i\]

which is the answer we wished to obtain.

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For those who want the AM-GM proof, here it is (I'm working backwards but you can see what you must do to get the identity):

\[a^2 + b^2 \geq 2ab\]

\[a^2+b^2 - ab \geq ab\]

\[(a+b)(a^2+b^2-ab) \geq ab(a+b)\]

\[a^3 + b^3 \geq a^2 b + ab^2\]

\[\dfrac {2}{3} a^3 + \dfrac {2}{3} b^3 \geq \dfrac {2}{3} a^2 b + \dfrac {2}{3} ab^2\]

\[a^3 + b^3 \geq \dfrac {1}{3} a^3 + \dfrac {1}{3} b^3 + \dfrac {2}{3} a^2 b + \dfrac {2}{3} ab^2\]

which is the identity. Thus proven.

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Cool!! Nice solution. But you should have waited some more time to post the solution. So that this note would have reached to many people.

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Sorry.

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Good Good

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