You visited us 10 times! Enjoying our articles? Unlock Full Access!

Cube Root of a Complex Number

If a2 + b2 ...

Question

If $a^{2} + b^{2} + c^{2} = 2, x^{2} + y^{2} + z^{2} = 2$ , where $a, b, c, x, y, z > 0$ , then maximum value of ax + by + cz is

2

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

\frac{3}{2}

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

\frac{1}{2}

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

None of these

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

Open in App

Solution

The correct option is A $2$
$A M \geq G M$
$\frac{a^{2} + x^{2}}{2} \geq \sqrt{a^{2} x^{2}}, \frac{b^{2} + y^{2}}{2} \geq \sqrt{b^{2} y^{2}}, \frac{c^{2} + z^{2}}{2} \geq \sqrt{c^{2} z^{2}}$
$\Rightarrow a^{2} + x^{2} \geq 2 a x, b^{2} + y^{2} \geq 2 b y, c^{2} + z^{2} \geq 2 c z$
$\Rightarrow a^{2} + x^{2} + b^{2} + y^{2} + c^{2} + z^{2} \geq 2 (a x + b y + c z)$
$\Rightarrow 2 (a x + b y + c z) \leq 2 + 2$
$\Rightarrow a x + b y + c z \leq 2$

Suggest Corrections

Similar questions

a^{2} + b^{2} + c^{2} = 1

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

a, b, c, x, y, z \geq 0,

a x + b y + c z

a^{2} + b^{2} + c^{2} = 1

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

a x + b y + c z \leq 1

a, b, c

x, y, z

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = 1, \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1, - \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

x = a^{2} - b c, y = b^{2} - c a, z = c^{2} + a b

\frac{(a + b + c) (x + y + z)}{a x + b y + c z}

a, b, c > R^{+}

a, b, c

x, y, z

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = 1

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

\frac{- x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

Join BYJU'S Learning Program