Zaltad wrote: Given such positive real numbers $a, b$ and $c$, that the system of equations: $ \{\begin{matrix}a^2x+b^2y+c^2z=1&&\\xy+yz+zx=1&&\end{matrix} $ has exactly one solution of real numbers $(x, y, z)$. Prove, that there is a triangle, which borders lengths are equal to $a, b$ and $c$. If $x+y=0$, then second equation implies $xy=1$, impossible. So $x+y\ne 0$ and second equation implies $z=\frac{1-xy}{x+y}$ Plugging in the first, we get $b^2y^2+y(b(a^2+b^2-c^2)x-1)+a^2x^2-x+c^2=0$ In order to have a unique solution, we need this quadratic in $y$ have à zero discriminant, and so : $((a^2+b^2-c^2)^2-4a^2b^2)x^2-2(a^2-b^2-c^2)x+1-4b^2c^2=0$ And this must be true for a unique value of $x$, which means that this quadratic in $x$ must also have a zero discriminant, which gives : $a^4+b^4+c^4-2a^2b^2-2b^2c^2-2a^2c^2=-1$ Which may be written $(a^2-(b+c)^2)(a^2-(b-c)^2)=-1$ which implies $|b-c|<a<|b+c|$ Hence the claim.