Factorization is an art (6)

Algebra Level 4

{ ( x y + y z + z x ) 2 = 3 ( x 2 y 2 + y 2 z 2 + z 2 x 2 ) x + y + z = 2016 x y z = w 3 \large \begin{cases} (xy+yz+zx)^2 & = 3(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}) \\ x+y+z & = 2016 \\ xyz & = w^{3} \end{cases}

Let x , y , z x, y, z and w w be positive real numbers satisfying the system of equations above, enter your answer as the value of w w .


The answer is 672.

Tommy Li by Tommy Li , 22, Hong Kong

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2 solutions

Ivan Koswara
Feb 17, 2017

By Cauchy-Schwarz inequality,

( ( x y ) 2 + ( x z ) 2 + ( y z ) 2 ) ⋅ ( 1 2 + 1 2 + 1 2 ) ≥ ( x y ⋅ 1 + x z ⋅ 1 + y z ⋅ 1 ) 2 ((xy)^2 + (xz)^2 + (yz)^2) \cdot (1^2 + 1^2 + 1^2) \ge (xy \cdot 1 + xz \cdot 1 + yz \cdot 1)^2

with equality if the vectors ( x y , x z , y z ) , ( 1 , 1 , 1 ) (xy, xz, yz), (1, 1, 1) are linearly dependent; in other words, x y = x z = y z xy = xz = yz , or x = y = z x = y = z . But we do achieve equality as given in the first equation. So x = y = z x = y = z . Plugging this to the second equation, we get x = y = z = 2016 3 = 672 x = y = z = \frac{2016}{3} = 672 , and plugging this to the third equation, we get w 3 = x y z = x 3 w^3 = xyz = x^3 , so w = x = 672 w = x = \boxed{672} (because w w is real, there's only one solution).

7 Helpful 0 Interesting 0 Brilliant 0 Confused
Tommy Li
Jul 11, 2016

( x y + y z + z x ) 2 = 3 ( x 2 y 2 + y 2 z 2 + z 2 x 2 ) (xy+yz+zx)^2 = 3(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2})

x 2 y 2 + y 2 z 2 + z 2 x 2 + 2 x 2 y z + 2 x y 2 z + 2 x y z 2 = 3 ( x 2 y 2 + y 2 z 2 + z 2 x 2 ) x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}+2x^2yz+2xy^2z+2xyz^2=3(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2})

2 x 2 y 2 + 2 y 2 z 2 + 2 z 2 x 2 − 2 x 2 y z − 2 x y 2 z − 2 x y z 2 = 0 2x^{2}y^{2}+2y^{2}z^{2}+2z^{2}x^{2}-2x^2yz-2xy^2z-2xyz^2=0

( x 2 y 2 − 2 x 2 y z + z 2 x 2 ) + ( x 2 y 2 − 2 x y 2 z + y 2 z 2 ) + ( y 2 z 2 − 2 x y z 2 + z 2 x 2 ) = 0 (x^2y^2-2x^2yz+z^2x^2)+(x^2y^2-2xy^2z+y^2z^2)+(y^2z^2-2xyz^2+z^2x^2) =0

x 2 ( y 2 − 2 y z + z 2 ) + y 2 ( x 2 − 2 x z + z 2 ) + z 2 ( y 2 − 2 y x + x 2 ) = 0 x^2(y^2-2yz+z^2)+y^2(x^2-2xz+z^2)+z^2(y^2-2yx+x^2)=0

x 2 ( y − z ) 2 + y 2 ( x − z ) 2 + z 2 ( y − x ) 2 = 0 x^2(y-z)^2+y^2(x-z)^2+z^2(y-x)^2=0

⇒ x 2 = y 2 = z 2 = 0 \Rightarrow x^2=y^2=z^2=0 (rej.)

⇒ y − z = x − z = y − x = 0 \Rightarrow y-z=x-z=y-x=0

⇒ x = y = z \Rightarrow x=y=z

x + y + z = 2016 x+y+z = 2016

⇒ x = y = z = 672 \Rightarrow x=y=z=672

x y z = 67 2 3 xyz=672^{3}

⇒ w = 672 \Rightarrow w=672

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Before reading the comments, can you spot the error in the solution? If so, how could it be fixed?

In 7th step: How do you conclude that x 2 = y 2 = z 2 = 0 x^2=y^2=z^2=0 ?

A Former Brilliant Member - 4 years, 11 months ago

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That's a good point. That presents some issues.

Calvin Lin Staff - 4 years, 3 months ago

Because the LHS is written as the sum of 3 squared numbers and it equal to 0, so each square has to equal to 0. x 2 ( y − z ) 2 = 0 ⇒ x 2 = 0 or ( y − z ) 2 = 0 x^2(y-z)^2=0\Rightarrow x^2=0\ \text{or}\ (y-z)^2=0

Hải Trung Lê - 4 years, 10 months ago

Unfortunately, the conclusion from x 2 ( y − z ) 2 + y 2 ( x − z ) 2 + z 2 ( y − x ) 2 = 0 x^2(y-z)^2+y^2(x-z)^2+z^2(y-x)^2=0 is not that x 2 = y 2 = z 2 = 0 x^2 = y^2 = z^2 = 0 or ( y − x ) = ( x − z ) = ( y − x ) = 0 (y-x) = (x-z) = ( y-x) = 0 .

Instead, it is just x ( y − z ) = y ( x − z ) = z ( y − x ) = 0 x(y-z) = y(x-z) = z (y-x) = 0 . There are then 8 possibilities to consider, and not just 2. Eg. we could have x = 0 , x − z = 0 , z = 0 x = 0, x - z = 0, z = 0 and y y is anything!

Calvin Lin Staff - 4 years, 3 months ago

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