A Good Triple

Algebra Level 5

A triple ( x , y , z ) ∈ C 3 (x, y, z) \in \mathbb{C^3} is called good iff the following conditions are satisfied. { ( x + y + z ) ( x 3 + y 3 + z 3 + x y z ) = x 2 ( x 2 − y 2 ) + y 2 ( y 2 − z 2 ) + z 2 ( z 2 − x 2 ) + 2014 2 x y z ( x y + y z + z x ) = 1007 \begin{cases} (x+y+z) \left( x^3 + y^3 + z^3 + xyz \right) & = x^2 \left( x^2- y^2 \right) + y^2 \left( y^2 - z^2 \right) + z^2 \left( z^2 - x^2 \right) + 2014\\ 2xyz \left( \sqrt{xy}+\sqrt{yz}+\sqrt{zx} \right) & = 1007 \end{cases} How many good triples consisting of positive reals are there?

This problem appeared in the Proofathon Algebra contest, and was posed by me.


The answer is 1.

Sreejato Bhattacharya by Sreejato Bhattacharya , 22, India

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1 solution

치원 윤
May 8, 2014

Notate ∑ s y m x a y b z c = ( a , b , c ) \sum_{sym} x^a y^b z^c = (a, b, c)

2 ( 3 , 1 , 0 ) + ( 2 , 1 , 1 ) + ( 2 , 2 , 0 ) = 4028 2(3, 1, 0)+(2, 1, 1)+(2, 2, 0)=4028

4 ( 3 2 , 3 2 , 1 ) = 4028 4(\frac{3}{2}, \frac{3}{2}, 1)=4028

By Muirhead's inequality,

2 ( 3 , 1 , 0 ) + ( 2 , 1 , 1 ) + ( 2 , 2 , 0 ) ≥ 4 ( 3 2 , 3 2 , 1 ) 2(3, 1, 0)+(2, 1, 1)+(2, 2, 0) ≥ 4(\frac{3}{2}, \frac{3}{2}, 1)

Equality holds, so x = y = z x=y=z

∴ \therefore There is only solution x = y = z = ( 1007 6 ) 1 4 x=y=z=(\frac{1007}{6})^\frac{1}{4}

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