$x^2+y^2+z^2=a$ then what is the of range of $xy+yz+zx$

Question

$$x^2+y^2+z^2=a$$ then what is the of range of $$xy+yz+zx$$ options

A) $[-a, a]$

B) $[-a/2, a/2]$

C) $[-a/2, a]$

Answer 1

since $$a=x^2+y^2+z^2\ge xy+yz+xz$$ this is true because $$(x-y)^2+(y-z)^2+(x-z)^2\ge 0$$ and the other hand,we have $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)\ge 0\Longrightarrow xy+yz+xz\ge-\dfrac{a}{2}$$ so $$xy+yz+xz\in[-\dfrac{a}{2},a]$$