If x, y, z are three + ive real numbers, then minimum value of y+zx+z+xy+x+yz is
A
1
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B
2
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C
3
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D
6
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Solution
The correct option is D 6 L. H. S =(yz+xy)+(zx+xz)+(yz+zy) Each pair is ≥2 as A.M. ≥G.M. ∴ L.H.S ≥6. Hence the minimum value is 6 Another form: yz(y+z−2x)+zx(z+x−2y)+xy(x+y−2z)≥0 or yz(y+z)+zx(z+x)+xy(x+y)≥6xyz or yx+zx+zy+xy+xz+yz>6 etc.