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Let a, b, c be nonzero real numbers such that a + b + c = 2 and a^2 + b^2 + c^2 = 4. Find \(\frac{ab}{c} + \frac{ac}{b} + \frac{bc}{a}.\)

 Sep 16, 2019
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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 

 

 

 

cool cool cool

 Sep 16, 2019
edited by CPhill  Sep 16, 2019

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