Question

We are giving the concept of arithmetic mean of mth power. Let a1,a2,a3,...an be n positive real numbers(not all equal) and m be a real number.Then am1+am2+am3+...+amnn≥(a1+a2+a3+...+ann)m if mεR∼[0.1]
However, if mε(0,1) then
am1+am2+am3+...+amnn<(a1+a2+a3+...+ann)m
Obviously, if m=0,1, then
am1+am2+am3+...+amnn=(a1+a2+a3+...+ann)m

If x>0,y>0,z>0x>0,y>0,z>0x>0,y>0,z>0x>0,y>0,z>0 is

• A. $0.2$
• B. $0.4$
• C. $0.6$
• D. $0.8$

Answer 1

The correct option is C $0.6$

Since, A.M of ${(-1)}^{th}$ powers$\ge {-1}^{th}$ powers of A.M
$\therefore ,\frac{{(2-x)}^{-1}+{(2-y)}^{-1}+{(2-z)}^{-1}}{3}\ge {\left(\frac{2-x+2-y+2-z}{3}\right)}^{-1}$
As $x+y+z=1$ we have as$={\left(\frac{6-x-y-z}{3}\right)}^{-1}={\left(\frac{6-1}{3}\right)}^{-1}=\frac{3}{5}$
or $\frac{1}{3}[\frac{1}{2-x}+\frac{1}{2-y}+\frac{1}{2-z}]\ge \frac{3}{5}$
or $\frac{1}{2-x}+\frac{1}{2-y}+\frac{1}{2-z}\ge \frac{9}{5}$
Multiply by $2$ we get $\frac{2}{2-x}+\frac{2}{2-y}+\frac{2}{2-z}\ge \frac{18}{5}$
Rearranging as $1+\frac{1}{2-x}+1+\frac{1}{2-y}+1+\frac{1}{2-z}\ge \frac{18}{5}$
By transposition,$\frac{x}{2-x}+\frac{y}{2-y}+\frac{z}{2-z}\ge \frac{18}{5}-3$
$\frac{x}{2-x}+\frac{y}{2-y}+\frac{z}{2-z}\ge \frac{3}{5}=0.6$
$\Rightarrow \frac{x}{2-x}+\frac{y}{2-y}+\frac{z}{2-z}\ge 0.6$
Thus, the minimum value of $\frac{x}{2-x}+\frac{y}{2-y}+\frac{z}{2-z}$ is $0.6$