If x-y-z -0-1 such that x - y - z - 1- then the least value of 1-x1-y1-zxyz is

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Properties of Determinants

If x,y,z ∈0,1...

Question

If $x, y, z \in (0, 1)$ such that $x + y + z = 1,$ then the least value of $\frac{(1 - x) (1 - y) (1 - z)}{x y z}$ is

8.0

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8.00

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Similar questions

x, y, z \in (0, 1)

x + y + z = 1,

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

x + y + z = 1

\frac{1}{x} + \frac{1}{y} + \frac{1}{z}

9

(1 - x) (1 - y) (1 - z) > 8 x y z

(1 - x) (1 - y) (1 - z) \geq k x y z

z = \frac{x^{1 / 25} - y^{1 / 25}}{x^{1 / 20} + y^{1 / 20}}

x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y}

	Column-I		Column-II
(A)	$a, b, c > 0$ then $\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b}$ can't take the value	(p)	$\frac{121}{81}$
(B)	p and q are positive real numbers and p + q = 1, then ${(p + \frac{1}{p})}^{2} + {(q + \frac{1}{q})}^{2}$ can't take the value	(q)	$\frac{343}{229}$
(C)	x, y, z are real numbers such that 0 <x, y, z< 1 and x + y + z = 2, then $\frac{x y z}{(1 - x) (1 - y) (1 - z)}$ can take the value	(r)	$\frac{133}{15}$
(D)	x, y, z are positive real numbers such that x + y + z = 1, then \frac{(1+x)(1+y)(1+z)}{(1-x)(1-y)(1-z)} can take the value	(s)	$\frac{413}{51}$
		(t)	8

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