who can solve it？

We have \[ \frac{AC_1}{AB}=2t=\frac{b \cos A}{c}=\frac{b^2+c^2-a^2}{2c^2} \] \[ \implies t=\frac{b^2+c^2-a^2}{4c^2} \] Substituting, \[ \left( \frac{t}{1-2t}\right)=\frac{\frac{b^2+c^2-a^2}{4c^2}}{\frac{2c^2-(b^2+c^2-a^2)}{2c^2}}=\frac{1}{2} \left(\frac{b^2+c^2-a^2}{c^2+a^2-b^2} \right)=\frac{1}{2} \left( \frac{b \cos A}{a \cos B} \right) \] \[ \implies \frac{a^2}{b^2} \left( \frac{t}{1-2t}\right)^2= \frac{1}{4} \left( \frac{\cos A}{\cos B} \right)^2 \] Analogously, we have \[ \frac{b^2}{c^2} \left( \frac{r}{1-2r}\right)^2=\frac{1}{4} \left( \frac{\cos B}{\cos C} \right)^2 \] \[ \frac{c^2}{a^2} \left( \frac{\mu}{1-2 \mu}\right)^2=\frac{1}{4} \left( \frac{\cos C}{\cos A} \right)^2 \] On the other hand, we get \[ 16tr \mu=\frac{1}{4} \left(\frac{(b^2+c^2-a^2)(c^2+a^2-b^2)(a^2+b^2-c^2)}{a^2b^2c^2} \right)=2 \cos A \cos B \cos C \] Therefore, the inequality reduces to \[ \left(\frac{\cos A}{\cos B} \right)^2 + \left(\frac{\cos B}{\cos C} \right)^2 + \left(\frac{\cos C}{\cos A} \right)^2 + 8 \cos A \cos B \cos C \ge 4 \] which is well known but I can't find a link, so a proof. Lemma: For reals $ x,y,z>0, xyz \le 1, $ we have $ \frac{x}{y}+\frac{y}{z}+\frac{z}{x} \ge x+y+z $. Applying the lemma to $ 4 \cos^2A, 4 \cos^2 B, 4 \cos^2 C $ (note that $ \prod \cos A \le \frac{1}{8} $), we get \[ \left(\frac{\cos A}{\cos B} \right)^2 + \left(\frac{\cos B}{\cos C} \right)^2 + \left(\frac{\cos C}{\cos A} \right )^2 \ge 4 \cos^2A+4 \cos^2 B+4 \cos^2 C \] therefore, \[ \left(\frac{\cos A}{\cos B} \right)^2 + \left(\frac{\cos B}{\cos C} \right)^2 + \left(\frac{\cos C}{\cos A} \right)^2 + 8 \cos A \cos B \cos C \] \[ \ge 4 \cos^2A+4 \cos^2 B+4 \cos^2 C + 8 \cos A \cos B \cos C = 4 \] because $ \cos^2 A+ \cos^2 B+ \cos^2 C+2 \cos A \cos B \cos C =1 $. $ \blacksquare $