If $x, y, z$ are three real numbers of the same sign, then the value of $(\frac{x}{y} + \frac{y}{z} + \frac{z}{x})$ lies in the interval

[2, \infty)

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[3, \infty)

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(3, \infty)

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(- \infty, 3)

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Solution

The correct option is B $[3, \infty)$
If we take $\frac{x}{y}, \frac{y}{z} . \frac{z}{x}$ as three real numbers, then

$A . M \geq G . M$ we get

$\frac{\frac{x}{y} + \frac{y}{z} + \frac{z}{x}}{3} \geq \sqrt[3]{\frac{x}{y} . \frac{y}{z} . \frac{z}{x}} \Rightarrow \frac{x}{y} + \frac{y}{z} + \frac{z}{x} \geq 3$

Hence $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} \in [3, \infty)$

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Similar questions

If x, y, z are three real numbers of the same sign then the value of x/y +y/z+z/x lies in the interval

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

x y + y z + z x

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

y z + z x + x y

x + y + z = 5

x y + y z + z x = 3

If $(x^{2} + y^{2} + z^{2}) = (x y + y z + z x)$ , $what is the value of x^{3} + y^{3} + z^{3}$ ?

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