2014/022) given

xyz = a/2..(1)

x^2+y^2 + z ^2 = a^2+ 6...(2)

x+y + z = a ...(3)
find 1/(xy + az) + 1/(yz + ax )+ 1/(zx + ay)

Solution

we have
xy + az = xy + z (x + y + z) = (z+x)(z+y) = (a-y)(a-z)
similarly
yz + ax = (a-y)(a-x)
zx + ay = (a-z)(a-x)

so 1/(xy + az) + 1/(yz + ax )+ 1/(zx + ay)
= 1/ (a-y)(a-z)+ 1/ (a-y)(a-x) + 1/ (a-z)(a-x)
= (3a – x – y – z)/)(a-y)(a-z)(a-x))


Numerator = 3a – a ( from (3)) = 2a

Denominator = a^3- (x+y+z)a^2 + (xy + yz + zx) a – xyz
= a^3- a. a^2 + (xy + yz + zx) a – xyz
= (xy + yz + zx) a – xyz

We have 2(xy + yz+ zx) = (x+y+z)^2 – (x^2 + y^2 + z^2) = a^2 – (a^2 + 6) = - 6

Or xy + yz + zx = - 3

So denominator = -3a – a/2 = - 7a/2

So value = - 4/7