China Mathematical Competition (Extra Test)

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 $w^2+1=(u+w)(v+w)$. $$\begin{array}{rl}\text{We get}\frac{{\cos }^{2}A}{1+\cos A}& =\frac{\frac{u^2}{u^2+1}}{1+\frac{u}{\sqrt{u^2+1}}}=\frac{u^2}{\sqrt{u^2+1}(\sqrt{u^2+1}+u)}\\ & =\frac{u^2(\sqrt{u^2+1}-u)}{\sqrt{u^2+1}}=u^2-\frac{u^3}{\sqrt{u^2+1}}\end{array}$$ $$\begin{array}{rl}& =u^2-\frac{u^3}{\sqrt{(u+v)(u+w)}}\\ & \ge u^2-\frac{u^3}{2}\left(\frac{1}{u+v}+\frac{1}{u+w}\right).\end{array}$$ 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 $\frac{{\cos }^{2}C}{1+\cos C}\ge w^2-\frac{w^3}{2}\left(\frac{1}{u+w}+\frac{1}{v+w}\right).$ \begin{aligned} \text{Hence}\quad f &\ge u^2 + v^2 + w^2 - \frac{1}{2} \left( \frac{u^3 + v^3}{u+v} + \frac{w^3 + v^3}{w+v} + \frac{u^3 + w^3}{u+w} \right) \\ &= u^2 + v^2 + w^2 - \frac{1}{2} \left[ (u^2 - uv + v^2) \\ & \qquad + (v^2 - vw + w^2) + (u^2 - uw^2 + w^2) \right] \\ &= \frac{1}{2} (uv + vw + uw) = \frac{1}{2}, \end{aligned} the equality sign is valid if and only if $u=v=w$, i.e. $a=b=c$, $x=y=z=\frac{1}{2}$, so $[f(x,y,z){]}_{\min }=\frac{1}{2}$.