Question

For set of three numbers $x,y,z,$ if $x^2+y^2+z^2=1$ and $A\le \Sigma xy\le B$, then set of values $[A,B]$ satisfying the condition

• 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 $[-\frac{1}{2},1]$

We know that $(x+y+z{)}^2\ge 0$

$x^2+y^2+z^2=1$

We have $x^2+y^2+z^2+2xy+2yz+2xz\ge 0$

$xy+yz+xz\ge -\frac{1}{2}$ ...(i)

Further $x^2+y^2+z^2+2xy+2yz+2xz=\frac{1}{2}\left[\right(x-y{)}^2+(y-z{)}^2+(z-x{)}^2]$

Using AM and GM we have

$x^2+y^2\ge 2xy{y}^2+z^2\ge 2yz{z}^2+x^2\ge 2zx$

By adding these three equations. We get,

$2\ge 2(xy+yz+xz)\dots \left(ii\right)$

$\therefore$ from $\left(i\right)$ and $\left(ii\right)$

$-\frac{1}{2}\le xy+yz+xz\le 1$