Let $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$ for x > 0, y > 0, z > 0 .Which of the following sets could be the possible values of (x - 1)(y - 1)(z - 1)?

[10,

\infty

)

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[11,

\infty

)

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[7,

\infty

)

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[6,

\infty

)

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Solution

The correct options are
A [11, $\infty$ )
B [10, $\infty$ )
$(\frac{1}{y} + \frac{1}{z}) (\frac{1}{z} + \frac{1}{x}) (\frac{1}{x} + \frac{1}{y}) \geq \frac{8}{x y z}$
From $A . M \geq G . M$
Hence $(1 - \frac{1}{x}) (1 - \frac{1}{y}) (1 - \frac{1}{z}) \geq \frac{8}{x y z}$
$\Rightarrow (x - 1) (y - 1) (z - 1) \geq 8$

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Let 1x+1y+1z=1 for x > 0, y > 0, z > 0 .Which of the following sets could be the possible values of (x - 1)(y - 1)(z - 1)?

The correct options are A [11, ∞) B [10, ∞)(1y+1z)(1z+1x)(1x+1y)≥8xyzFrom A.M≥G.MHence (1−1x)(1−1y)(1−1z)≥8xyz⇒(x−1)(y−1)(z−1)≥8

Let $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$ for x > 0, y > 0, z > 0 .Which of the following sets could be the possible values of (x - 1)(y - 1)(z - 1)?