Olympiad Proof Problem - Day 6

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 x1x_{1}, x2x_{2}, x3x_{3}, \ldots xnx_{n} are positive real numbers, then prove the following inequality

i=1nxi3xi2+xixi+1+xi+12x1+x2+x3++xn3\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:

  • xn+1=x1x_{n+1} = x_{1}.

Try more proof problems at Olympiad Proof Problems.

Note by Surya Prakash
3 years, 11 months ago

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Observe that

0=(x1x2)+(x2x3)++(xnx1)=i=1nxi3xi+13xi2+xixi+1+xi+120 = (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,

i=1nxi3xi2+xixi+1+xi+12=12i=1nxi3+xi+13xi2+xixi+1+xi+12\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 a3+b313a3+23a2b+23ab2+13b3=13(a+b)(a2+ab+b2)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:

12i=1nxi3+xi+13xi2+xixi+1+xi+1212i=1nx1+xi+13=13i=1nxi\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.

Sharky Kesa - 3 years, 11 months ago

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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.

Surya Prakash - 3 years, 11 months ago

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Sorry.

Sharky Kesa - 3 years, 11 months ago

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@Sharky Kesa It's ok.

Surya Prakash - 3 years, 11 months ago

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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):

a2+b22aba^2 + b^2 \geq 2ab

a2+b2ababa^2+b^2 - ab \geq ab

(a+b)(a2+b2ab)ab(a+b)(a+b)(a^2+b^2-ab) \geq ab(a+b)

a3+b3a2b+ab2a^3 + b^3 \geq a^2 b + ab^2

23a3+23b323a2b+23ab2\dfrac {2}{3} a^3 + \dfrac {2}{3} b^3 \geq \dfrac {2}{3} a^2 b + \dfrac {2}{3} ab^2

a3+b313a3+13b3+23a2b+23ab2a^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.

Sharky Kesa - 3 years, 11 months ago

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Good Good

Ayanlaja Adebola - 3 years, 11 months ago

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