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:**

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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

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.

Log in to reply

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.

Log in to reply

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.

Log in to reply

Sorry.

Log in to reply

Log in to reply

Good Good

Log in to reply