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If x,y,z are positive real numbers such ...

If `x,y,z` are positive real numbers such that `x^(2)+y^(2)+Z^(2)=7` and `xy+yz+xz=4` then the minimum value of `xy` is

A

`1`

B

`(1)/(2)`

C

`(1)/(4)`

D

`(1)/(8)`

Text Solution

Verified by Experts

The correct Answer is:
C
`(c )` `xy=4-(x+y)z`
Now consider `x+y` and `z`
Using `G.M. ge A.M.`, we have
`sqrt((x+y)z) le (x+y+z)/(2)`
Also, `(x+y+z)^(2)`
`=x^(2)+y^(2)+z^(2)+2(xy+yz+zx)`
`=7+8=15`
Now `(x+y)z le ((x+y+z)^(2))/(4) le (15)/(4)`
`implies (x+y)z|_(max)=(15)/(4)`
`impliesxy|_(min)=4-(15)/(4)=(1)/(4)`
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