Bunuel wrote:
If \(x + y + z < 1\), is \(z < −1\)?
(1) x and y are positive numbers.
(2) xy = 1.
x + y + z < 1
(1) x & y are positive
Assume x=1/4, y= 1/3, z=1/3 here z is not less than -1
Now Assume x = 1/4, y=1/3, z=-3 here z is less than -1
Hence Not Sufficient
(2) xy = 1
Assume x = -1, y=-2, z=-3, here z is less than -1
Now Assume x=-1, y=-1, z=3/2 here z is not less than -1
Hence not sufficient
On combining
Since x and y are positive and xy=1
x=1, y=1 then z has to be <-1
if x=1/3, y=3 then z has to be <-1
if x = 3/4 y = 4/3 then also z has to be <-1
Also
since xy=1 we can write x = 1/y and then
x + y + z < 1
1/y + y + z < 1
z<1-y-1/y
z<1-(y+1/y) (we know x+1/x>=2 if x>0)
taking minimum value of y+1/y
z<1-2
z<-1
Hence C
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