system of equations solving for positive a , b , c ( 1 − a ) ( 1 − b ) ( 1 − c ) = a b c a + b + c = 1I found that 0 &lt; a , b , c &lt; 1 and I try to solve it by try ( 1 − a ) = a, ( 1 − b ) = b and ( 1 − c ) = c and got that a = 0.5 , b = 0.5 , c = 0.5 but it contradicts the second equations.

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system of equations solving for positive   a  ,  b  ,  c  (  1  −  a  )  (  1  −  b  )  (  1  −  c  )  =  a  b  c  a  +  b  +  c  =  1I found that   0  &amp;lt;  a  ,  b  ,  c  &amp;lt;  1 and I try to solve it by try   (  1  −  a  )  =  a,   (  1  −  b  )  =  b and   (  1  −  c  )  =  c and got that   a  =  0.5  ,  b  =  0.5  ,  c  =  0.5 but it contradicts the second equations.

aqueritztv · Accepted Answer

by expanding the left hand side you will have  1  −  (  a  +  b  +  c  )  +  a  b  +  b  c  +  a  c  −  a  b  c  =  a  b  cusing the second equality we will have:  2  a  b  c  =  a  b  +  b  c  +  c  aAnd using the substitution method we can replace   a by   1  −  (  b  +  c  ) then we have:  2  b  c  (  1  −  (  b  +  c  )  )  =  (  1  −  (  b  +  c  )  )  (  b  +  c  )  +  b  c  ⇒  2  b  c  −  2      b    2    c  −  2  b      c    2    =  b  +  c  −      b    2    −      c    2    −  2  b  c  +  b  c  ⇒  2      b    2    c  +  2  b      c    2    +  b  +  c  −      b    2    −      c    2    −  3  b  c  =  0    s  .  t  .    0  ≤  b  ,  c  ≤  1

Zeihergp · Accepted Answer

From   (  1  −  a  )  (  1  −  b  )  (  1  −  c  )  =  a  b  c you cannot deduce   (  1  −  a  )  =  1 and so on. One set of solutions is to make both sides zero-let   a  =  1  ,  b  =  0  ,  c  =  0 or any permutation. The symmetry means that any permutation of a solution will also be a solution. Given one cubic and one linear equation you expect three solutions and we have that many.

system of equations solving for positive a , b , c ( 1 −<!-- − -

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