NMTC 2017 sub junior

To Find $S = a^2 + b^2 + c^2 + 2abc$ $x^2(y+z)=a^2 -----(1)$ $y^2(z+x)=b^2 -----(2)$ $z^2(x+y)=c^2 -----(3)$ $xyz=abc -----(4)$ Multiplying (1),(2)and (3)

$x^2y^2z^2(x + y)(y + z)(z + x) = a^2b^2c^2$ $(x + y)(y + z)(z + x) = 1 -------(5)$ $S = a^2 + b^2 + c^2 + 2abc = x^2(y + z) + y^2(z + x) + z^2(x + y) + 2xyz -----(6)$ Put $y+z=0$ $S= 0 + z^2(z + x) + z^2(x - z) + 2x(-z)(z)$ $S= z^3 + z^2x + z^2x - z^3 - 2xz^2$ $S= 0$ $\therefore (y + z)$ is a factor

similarily $(x + y)$ & $(x + z)$ are also factors

$x^2(y + z) + y^2(z + x) + z^2(x + y) + 2xyz = k(x + y)(y + z)(z + x) -----(7)$

put $x = 0, y = 1, z = 1$ we get $k = 1$ From above Equation (7)

Now From (6) and (7) $S = k(x + y)(y + z)(z + x)$ $\implies \quad \boxed{S = 1}$