Question
If $${x^2} + {y^2} + {z^2} = 1$$ then the value of $$xy + yz + zx$$ lies in the interval
A.
$$\left[ {\frac{1}{2},2} \right]$$
B.
$$[ - 1, 2]$$
C.
$$\left[ - {\frac{1}{2},1} \right]$$
D.
$$\left[ { - 1,\frac{1}{2}} \right]$$
Answer :
$$\left[ - {\frac{1}{2},1} \right]$$
Solution :
Let $$xy + yz + zx = \lambda .$$ Then
$$\eqalign{
& {x^2} + {y^2} + {z^2} - \lambda = \frac{1}{2}\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right] \geqslant 0 \cr
& \Rightarrow \,\,1 - \lambda \geqslant 0. \cr
& {\text{Again, }}{\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2\lambda = 1 + 2\lambda \cr
& \Rightarrow \,\,1 + 2\lambda \geqslant 0. \cr} $$