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  4. If x, y, z>0 and x+y+z=1, then find the least value of E=(2 x/1-x)+(2 y/1-y)+(2 z/1-z).

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Q. If $x, y, z>0$ and $x+y+z=1$, then find the least value of $E=\frac{2 x}{1-x}+\frac{2 y}{1-y}+\frac{2 z}{1-z}$.

Sequences and Series

Solution:

$E=2\left[\frac{x}{1-x}+\frac{y}{1-y}+\frac{z}{1-z}\right]=2\left[\frac{1}{1-x}+\frac{1}{1-y}+\frac{1}{1-z}-3\right]$
Now, $ \frac{(1-x)+(1-y)+(1-z)}{3} \geq \frac{3}{\frac{1}{1-x}+\frac{1}{1-y}+\frac{1}{1-z}}$
$\frac{2}{3} \geq \frac{3}{\frac{1}{1-x}+\frac{1}{1-y}+\frac{1}{1-z}}$
$\Rightarrow \frac{1}{1-x}+\frac{1}{1-y}+\frac{1}{1-z} \geq \frac{9}{2}$
$\Rightarrow E \geq 2\left[\frac{9}{2}-3\right]=3 \Rightarrow E _{\min }=3$