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Questions  

 If x,y and z are real and different and u=x2+4y2+9z26yz3zx2xy, then u is always 

a
non negative
b
zero
c
non positive
d
none of these

detailed solution

Correct option is A

u=x2+4y2+9z2−6yz−3zx−2xy=122x2+8y2+18z2−12yz−6zx−4xy=12x2−4xy+4y2+4y2+9z2−12yz+x2+9z2−6zx=12(x−2y)2+(2y−3z)2+(3z−x)2≥0 Thus, u is always non-negative.

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