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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

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\frac{3}{2}

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\frac{1}{2}

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None of these

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Solution

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

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