If-x-y-z-1-and-x-y-z-are-positive-real-numbers-then-the-least-value-of-1-x-1-1-y-1-1-z-1-is-

Question Number 112998 by Aina Samuel Temidayo last updated on 10/Sep/20

If x + y + z = 1 and x, y, z are positive

real numbers, then the least value of

(\frac{1}{x} - 1) (\frac{1}{y} - 1) (\frac{1}{z} - 1) is

Answered by MJS_new last updated on 11/Sep/20

z = 1 - x - y

\Rightarrow - \frac{(x - 1) (x + y) (y - 1)}{x (x + y - 1) y}

\frac{d}{d y} [- \frac{(x - 1) (x + y) (y - 1)}{x (x + y - 1) y}] = 0

- \frac{(x - 1) (x + 2 y - 1)}{{(x + y - 1)}^{2} y^{2}} = 0 \Rightarrow y = \frac{1 - x}{2}

\Rightarrow - \frac{{(x + 1)}^{2}}{x (x - 1)}

\frac{d}{d x} [- \frac{{(x + 1)}^{2}}{x (x - 1)}] = 0

\frac{(x + 1) (3 x - 1)}{x^{2} {(x - 1)}^{2}} = 0 \Rightarrow x = y = z = \frac{1}{3}

\Rightarrow minimum value is 8

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