Let a, b, c be nonzero real numbers such that a + b + c = 2 and a^2 + b^2 + c^2 = 4. Find \(\frac{ab}{c} + \frac{ac}{b} + \frac{bc}{a}.\)
+128475 a + b + c = 2 (1) square both sides ⇒ (a^2 + b^2 + c^^2) + 2(ab + ac + bc) = 4 (2)
a^2 + b^2 + c^2 = 4 (3)
Sub (3) into (2) and we have that
(4) + 2(ab + ac + bc) = 4
2(ab + ac + bc) = 0
ab + ac + bc = 0
(ab + ac + bc)^2 = 0 square both sides
(ab)^2 + (ac)^2 + (bc)^2 + 2[a^2bc + ab^2c + abc^2] = 0 ⇒
(ab)^2 + (ac)^2 + (bc)^2 + 2abc ( a + b + c) = 0 ⇒
(ab)^2 + (ac)^2 + (bc)^2 = -2abc ( a + b + c) (4)
Find ab ac bc
___ + __ + ___ =
c b a
ab*ab + ac*ac + bc*bc
___________________ =
abc
(ab)^2 + (ac)^2 + (bc)^2
___________________ sub ( 4) for the numerator and we have that
abc
-2abc ( a + b + c)
_______________ =
abc
-2 (a + b + c) sub (1) into this
-2 (2) =
-4
