If $x^{2} + y^{2} + z^{2} = 1$ then the value of $x y + y z + z x$ lies in the interval

[\frac{1}{2}, 2]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[- 1, 2]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[- \frac{1}{2}, 1]

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

[- 1, \frac{1}{2}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $[- \frac{1}{2}, 1]$
$(x + y + z)^{2} \geq 0$
$\Rightarrow x y + y z + z x \geq - 1 / 2$
Now
$A . M \geq G . M$
$(x y + y z + z x) / 3 \geq \sqrt[3]{x^{2} y^{2} z^{2}}$
$(x^{2} + y^{2} + z^{2}) / 3 \geq \sqrt[3]{x^{2} y^{2} z^{2}}$
$\Rightarrow (x y + y z + z x) / 3 \leq 1 / 3$
So $x y + y z + z x ϵ [- 1 / 2, 1]$

Suggest Corrections

Similar questions

x^{2} + y^{2} + z^{2} = 1

y z + z x + x y

y^{2} + y z + z^{2} = 49 . . . . . (1)

z^{2} + z x + x^{2} = 19 . . . . (2)

x^{2} + x y + y^{2} = 39 . . . . (3)

x^{2} + y^{2} + z^{2} - x y - y z - z x = \frac{1}{2} [(x - y)^{2} + (y - z)^{2} + (z - x)^{2}]

x y + y z + z x = 1

\frac{x}{1 + x^{2}} + \frac{y}{1 + y^{2}} + \frac{z}{1 + z^{2}}

x^{2} + y^{2} + z^{2} = 1

Join BYJU'S Learning Program