algebric expression

quantskillsgmat wrote:x,y and z are real numbers such that x+y+z=5 and xy+yz+zx is 3. what can be minimum value of x.
a)5/3
b)13/3
c)1/3
d)none

x + y = 5 - z
(x + y)² = (5 - z)²
xy = 3 - z(y + x)
xy = 3 - z(5 - z)
Now (x - y)² = (x + y)² - 4xy and (x - y)² ≥ 0
So, (x + y)² - 4xy ≥ 0
(5 - z)² - 4[3 - z(5 - z] ≥ 0
25 + z² - 10z - 12 + 20z - 4z² ≥ 0
-3z² + 10z + 13 ≥ 0
-(z + 1)(3z - 13) ≥ 0
-1 ≤ z ≤ 13/3
The maximum possible value of z is 13/3. z can be maximum if x and y are minimum. This means minimum value of x = y = 1/3, as 1/3 + 1/3 + 13/3 = 5 and 1/9 + 13/9 + 13/9 = 3.

The correct answer is C.