If x, y and z are positive real numbers such that $x + y + z = a$ then-

\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{a}

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\frac{1}{x} + \frac{1}{y} + \frac{1}{z} < \frac{9}{a}

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(a - x) (a - y) (a - z) > \frac{8}{27} a^{3}

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(a - x) (a - y) (a - z) > a^{3}

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Solution

The correct option is A $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{a}$
Since $A M \geq H M$
$\frac{x + y + z}{3} \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}$
$\Rightarrow \frac{a}{3} \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}$ or $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{a}$

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