Quadratic Diophantines!

Relevant wiki: Lagrange Multipliers

We treat the first equation as a constraint only, By AM-GM $\displaystyle 27 xyz(x+y+z)^3 \le (x+y+z)^6$ but $3(x^2+y^2+z^2)\ge (x+y+z)^2$ . Since we are given that $3(x^2+y^2+z^2)=1$ we claim $(x+y+z)^6\le 1$ .

So, $\displaystyle xyz(x+y+z)^3\le \frac{1}{27}$ . Now consider the expression $\displaystyle \frac{x^2 y^2+y^2 z^2+z^2 x^2}{xyz(x+y+z)^3}\ge 27(x^2 y^2+y^2 z^2+z^2 x^2)$

We need to minimize this with the constraint $3(x^2+y^2+z^2)=1$ . Set $x^2\equiv a,y^2\equiv b,z^2\equiv c$

If we define the lagrangian as $L_\lambda(a,b,c)=ab+bc+ca+\lambda(3(a+b+c)-1)$

Solving the partial derivatives gives $\displaystyle a=b=c=\frac{1}{9}$ . And by using Hessian matrix we are sure that it's a minimum. The minimum is $1$

So, $\displaystyle \frac{x^2 y^2+y^2 z^2+z^2 x^2}{xyz(x+y+z)^3}\ge 27(x^2 y^2+y^2 z^2+z^2 x^2)\ge 1$ but here have $\displaystyle \frac{x^2 y^2+y^2 z^2+z^2 x^2}{xyz(x+y+z)^3}\ge 27(x^2 y^2+y^2 z^2+z^2 x^2)=1$ so equality holds and we must have,

$\displaystyle x=y=z=\pm\frac{1}{3}$ and solutions.