Question

If $x^2+y^2+z^2=1$, then $yz+zx+xy$ lies in the interval

• A. $[\frac{1}{2},2]$
• B. $[-1,2]$
• C. $[-\frac{1}{2},1]$
• D. $[-1,\frac{1}{2}]$

Answer 1

The correct option is C $[-\frac{1}{2},1]$

$x^2+y^2+z^2=1\dots \left[1\right]$
$(x-y{)}^2+(y-z{)}^2+(z-x{)}^2=2(x^2+y^2+z^2-xy-yz-zx)$
$(x-y{)}^2+(y-z{)}^2+(z-x{)}^2=2[1-(xy+yz+zx)]$
$xy+yz+zx=1-\frac{1}{2}\left[\right(x-y{)}^2+(y-z{)}^2+(z-x{)}^2]$
$xy+yz+zx\le 1\dots \left[2\right]$
As $\left[\right(x-y{)}^2+(y-z{)}^2+(z-x{)}^2\ge 0]$

Again,
$(x+y+z{)}^2=x^2+y^2+z^2+2(xy+yz+zx)$
$\Rightarrow (x+y+z{)}^2=1+2(xy+yz+zx)$
$\Rightarrow xy+yz+zx=\frac{1}{2}\left[\right(x+y+z{)}^2-1]$
$\Rightarrow xy+yz+zx\ge -\frac{1}{2}\dots \left[3\right]$
$[\because (x+y+z{)}^2\ge 0]$

From $\left[2\right]$ and $\left[3\right]$,
$xy+yz+zx\in [-\frac{1}{2},1]$