x+y+z=0
y+z = -x
(y+z)(y-z)= -x(y-z)
y² -z² = -x(y -z)
x(y² - z²) = -x² ( y - z)
x(z² - y² ) = x² ( y -z)
x(z² -y² ) + x² ( y -z) = x² (y-z) + x² (y - z)
x( z² - y² + xy - xz ) = 2x² ( y - z)
x [ z(z-x) + y(x - y) ] = 2x² (y -z)
x [ z(z-x) + y(x - y) ] / ( yz) = 2x² (y -z) / (yz)
x[ (z-x) /y + (x-y) /z ] = 2x² (y -z) / (yz)
(x/ (y-z)) [ (z-x) /y + (x-y) /z ] = 2x² / (yz)
posons a=x/(y-z) b=y/(z-x) c=z/(x-y)
on vient de démontrer que
a( 1/b + 1/c) = 2x² /(yz) mais on aurait pu aussi montrer que
b(1/c+1/a)=2y²/(zx)) et
c(1/a+1/b)=2z²/(xy)
A=(a+b+c) (1/a+1/b+1/c)= 1 +a(1/b+1/c) + 1+b(1/c+1/a) + 1+c(1/a+1/b)=
3 + 2(x² /yz + y²/zx + z²/xy) = 3 + 2(x^3 +y^3+z^3)/(xyz)
(x+y+z)^3 =x^3+y^3+z^3+3x²y+3y²z+3z²x +3xy²+3yz²+3zx²+6xyz= 0
3x²y+3xy² =3xy(y+x)= -3xyz
3y²z+3yz² = 3yz(y+z)=-3xyz
3z²x+3zx²=-3yz
x^3 +y^3+z^3 -9xyz+6xyz= 0
x^3+y^3+z^3 = 3xyz
A=3+2(3)xyz/(xyz)= 3 +6 = 9
