Problem 1664

By assumption, $b(az + cx - b) + c(bx + ay - c) - a(cy + bz - a) = 0$, i.e. $2bcx + a^2 - b^2 - c^2 = 0$, we get $x = \frac{b^2 + c^2 - a^2}{2bc}$. For the similar reason, $y = \frac{a^2 + c^2 - b^2}{2ac}$ and $z = \frac{a^2 + b^2 - c^2}{2ab}$.

Since $a$, $b$, $c$, $x$, $y$, $z$ are positive, by the above three expressions, we know $b^2 + c^2 > a^2$, $a^2 + c^2 > b^2$ and $a^2 + b^2 > c^2$. Thus there is an acute triangle $ABC$ with the lengths of its sides $a$, $b$, $c$. So $x = \cos A$, $y = \cos B$ and $z = \cos C$. The problem is now changed to finding the minimum value of the function
$$f(\cos A, \cos B, \cos C) = \frac{\cos^2 A}{1+\cos A} + \frac{\cos^2 B}{1+\cos B} + \frac{\cos^2 C}{1+\cos C}$$
Set $u = \cot A$, $v = \cot B$, $w = \cot C$, then $u, v, w \in \mathbb{R}^+$, $uv + vw + wu = 1$, $u^2 + 1 = (u+v)(u+w)$, $v^2 + 1 = (u+v)(v+w)$ and $w^2 + 1 = (u+w)(v+w)$.
$$\begin{aligned} \text{We get}\quad \frac{\cos^2 A}{1+\cos A} &= \frac{\frac{u^2}{u^2+1}}{1+\frac{u}{\sqrt{u^2+1}}} = \frac{u^2}{\sqrt{u^2+1}(\sqrt{u^2+1}+u)} \\ &= \frac{u^2(\sqrt{u^2+1}-u)}{\sqrt{u^2+1}} = u^2 - \frac{u^3}{\sqrt{u^2+1}} \end{aligned}$$
$$\begin{aligned} &= u^2 - \frac{u^3}{\sqrt{(u+v)(u+w)}} \\ &\ge u^2 - \frac{u^3}{2} \left( \frac{1}{u+v} + \frac{1}{u+w} \right). \end{aligned}$$
By a similar argument, $\frac{\cos^2 B}{1 + \cos B} \ge v^2 - \frac{v^3}{2} \left( \frac{1}{u+v} + \frac{1}{v+w} \right)$
and
$\frac{\cos^2 C}{1 + \cos C} \ge w^2 - \frac{w^3}{2} \left( \frac{1}{u+w} + \frac{1}{v+w} \right).$
$$\text{Hence f u 2 + v 2 + w 2 - 1 2 ( u 3 + v 3 u+v + w 3 + v 3 w+v + u 3 + w 3 u+w ) = u 2 + v 2 + w 2 - 1 2 [ (u 2 - uv + v 2) + (v 2 - vw + w 2) + (u 2 - uw 2 + w 2) ] = 1 2 (uv + vw + uw) = 1 2 ,}$$
the equality sign is valid if and only if $u = v = w$, i.e. $a = b = c$, $x = y = z = \frac{1}{2}$,
so $[f(x, y, z)]_{\min} = \frac{1}{2}$.