Find the expression value

Algebra Level pending

Positive real numbers x x , y y , and z z are such that x + y + z + x y z = 4 x+y+z+\sqrt{xyz} =4 . Find value of the expression:

( 4 − x ) ( 4 − y ) z + ( 4 − y ) ( 4 − z ) x + ( 4 − z ) ( 4 − x ) y − x y z \sqrt{(4-x)(4-y)z} + \sqrt{(4-y)(4-z)x} + \sqrt{(4-z)(4-x)y} - \sqrt{xyz}


The answer is 8.

Danish Ahmed by Danish Ahmed , 24, India

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1 solution

Chew-Seong Cheong
Nov 28, 2020

Consider the first radical:

( 4 − x ) ( 4 − y ) z = ( 16 − 4 ( x + y ) + x y ) z = ( 16 − 4 ( x + y + z ) + 4 z + x y ) z = ( 16 − 4 ( 4 − x y z ) + 4 z + x y ) z = ( 4 x y z + 4 z + x y ) z = ( ( 2 z ) 2 + 2 ( 2 z ) x y + ( x y ) 2 ) z = ( 2 z + x y ) 2 z = 2 z + x y z \begin{aligned} \sqrt{(4-x)(4-y)z} & = \sqrt{(16-4(x+y) + xy)z} \\ & = \sqrt{(16-4(x+y+z) + 4z + xy)z} \\ & = \sqrt{(16-4(4-\sqrt{xyz}) + 4z + xy)z} \\ & = \sqrt{(4\sqrt{xyz} + 4z + xy)z} \\ & = \sqrt{((2\sqrt z)^2+2(2\sqrt z) \sqrt{xy}+ (\sqrt{xy})^2)z} \\ & = \sqrt{(2\sqrt z + \sqrt{xy})^2z} \\ & = 2 z + \sqrt{xyz} \end{aligned}

Therefore we have:

S = ( 4 − x ) ( 4 − y ) z + ( 4 − y ) ( 4 − z ) x + ( 4 − z ) ( 4 − x ) y − x y z = 2 z + x y z + 2 x + x y z + 2 y + x y z − x y z = 2 ( x + y + z + x y z ) = 8 \begin{aligned} S & = \sqrt{(4-x)(4-y)z} + \sqrt{(4-y)(4-z)x} + \sqrt{(4-z)(4-x)y} - \sqrt{xyz} \\ & = 2z + \sqrt{xyz} + 2x + \sqrt{xyz} + 2y + \sqrt{xyz} - \sqrt{xyz} \\ & = 2(x+y+z + \sqrt{xyz}) \\ & = \boxed 8 \end{aligned}

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