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If x > 0, y > 0, z>0 and x + y + z = 1 t...

If `x > 0, y > 0, z>0 and x + y + z = 1` then the minimum value of `x/(2-x)+y/(2-y)+z/(2-z)` is:

A

(a) `0.2`

B

(b) `0.4`

C

(c) `0.6`

D

(d) `0.8`

Text Solution

Verified by Experts

The correct Answer is:
C
Since, AM of `(-1)th` powers `ge(-1)th` powers of AM
`:.((2-x)^(-1)+(2-y)^(-1)+(2-z)^(-1))/(3)ge(((2-x)+(2-y)+(2-z))/(3))^(-1)`
`=[(6-(x+y+z))/(3)]^(-1)=((6-1)/(3))^(-1)=(3)/(5)" " [:.x+y+z=1]`
`((2-x)^(-1)+(2-y)^(-1)+(2-z)^(-1))/(3)ge (3)/(5)`
or `(1)/(3)[(1)/(2-x)+(1)/(2-y)+(1)/(2-z)]ge (3)/(5)`
`implies (1)/(2-x)+(1)/(2-y)+(1)/(2-z)ge (9)/(5)`
or `(2)/(2-x)+(2)/(2-y)+(2)/(2-z)ge (18)/(5)`
or `1+(x)/(2-x)+1+(y)/(2-y)+1+(z)/(2-z)ge (18)/(5)`
or `(x)/(2-x)+(y)/(2-y)+(z)/(2-z)ge (18)/(5)-3`
Hence, `(x)/(2-x)+(y)/(2-y)+(z)/(2-z)ge (3)/(5)=0.6`
`implies (x)/(2-x)+(y)/(2-y)+(z)/(2-z)ge 0.6`
Thus, minimum value of `(x)/(2-x)+(y)/(2-y)+(z)/(2-z)` is 0.6.
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