Question

$x,y,z$ are positive real numbers and $x+y+z=\frac{1}{2}$ then

• A. $\left(\frac{1}{2}\right)<\left(1-x\right)\left(1-y\right)\left(1-z\right)<\left(\frac{1}{3}\right)$
• B. $\left(\frac{1}{2}\right)<\left(1-x\right)\left(1-y\right)\left(1-z\right)<\left(\frac{2}{3}\right)$
• C. $\left(\frac{1}{2}\right)<\left(1-x\right)\left(1-y\right)\left(1-z\right)<1$
• D. None of these

Answer 1

The correct option is B

$\left(\frac{1}{2}\right)<\left(1-x\right)\left(1-y\right)\left(1-z\right)<\left(\frac{2}{3}\right)$

Explanation for the correct option:

Find the interval in which the value of the expression $\left(1-x\right)\left(1-y\right)\left(1-z\right)$ lie

We know that $\left(1–y_n\right)<(1–x_1)(1–x_2)\dots \dots (1–x_n)<\left(\frac{1}{1+y_n}\right)$ , where $y_n=x_1+x_2+....+x_n$

It is given that,

$x,y,z\in R^{+}$ and $x+y+z=\frac{1}{2}$

Using the above inequality, we get

$\left(1-\left(x+y+z\right)\right)<\left(1-x\right)\left(1-y\right)\left(1-z\right)<\left(\frac{1}{1+\left(x+y+z\right)}\right)$

$\Rightarrow$ $\left(1-\frac{1}{2}\right)<\left(1-x\right)\left(1-y\right)\left(1-z\right)<\left(\frac{1}{1+{\displaystyle \frac{1}{2}}}\right)$

$\Rightarrow$ $\left(\frac{2-1}{2}\right)<\left(1-x\right)\left(1-y\right)\left(1-z\right)<\left(\frac{1}{{\displaystyle \frac{2+1}{2}}}\right)$

$\Rightarrow$ $\left(\frac{1}{2}\right)<\left(1-x\right)\left(1-y\right)\left(1-z\right)<\left(\frac{1}{{\displaystyle \frac{3}{2}}}\right)$

$\Rightarrow$ $\left(\frac{1}{2}\right)<\left(1-x\right)\left(1-y\right)\left(1-z\right)<\left(\frac{2}{{\displaystyle 3}}\right)$

Hence, the correct option is $B)\left(\frac{1}{2}\right)<\left(1-x\right)\left(1-y\right)\left(1-z\right)<\left(\frac{2}{{\displaystyle 3}}\right)$.