Vietnamese Mathematical Competitions

Putting $m = a^2$, $n = b^2$, $p = c^2$ then $m, n, p > 0$. We have the system
$$\begin{cases} mx + ny + pz = 1 \\ xy + yz + zx = 1 \end{cases}.$$
From this system, we get $mx = 1 - ny - pz$ and $m(xy + yz + zx) = m$, therefore
$$ny^2 + y(nz + pz - mz - 1) + m - z + pz^2 = 0 \quad (*)$$
Equation $(*)$ has only one solution when $(nz + pz - mz - 1)^2 - 4n(m - z + pz^2) = 0$.
This equation is equivalent to $((m - n - p)^2 - 4np)z^2 + 2(m + n - p)z + 1 - 4mn = 0$. $(**)$
Equation $(**)$ has only one solution when
$$(m + n - p)^2 = (1 - 4mn)((m - n - p)^2 - 4np).$$
or by simple algebra $m^2 + n^2 + p^2 + 1 = 2(mn + np + pm)$.
Conversely, with this condition, putting back into $(*)$ and $(**)$, we see that the system has only one solution: $(x, y, z) = (n + p - m, p + m - n, m + n - p)$.
Thus, $m^2 + n^2 + p^2 + 1 = 2(mn + np + pm) \Leftrightarrow a^4 + b^4 + c^4 + 1 = 2(a^2b^2 + b^2c^2 + c^2a^2)$ is the necessary and sufficient condition for the system to have only one solution.
Moreover, from this condition, we have
$$2(a^2b^2 + b^2c^2 + c^2a^2) > a^4 + b^4 + c^4 \Leftrightarrow (a + b + c)(a + b - c)(b + c - a)(c + a - b) > 0.$$
From which follows that all of the numbers $a + b - c$, $b + c - a$, $c + a - b$ are positive, so $a, b, c$ are the sides of certain triangle. (Q.E.D)