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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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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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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Good Good

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