Geometry in inequality part II

Let $x=tan A,y=tan B,c=tan C$ From the conditional we have: $x=\frac{y-z}{1+yz}$ $\Leftrightarrow tanA=\frac{tanB-tanC}{1+tanB\times tanC}=tan(B-C)\Leftrightarrow A=B-C+k\pi$ So $\frac{-\pi }{2}< A-B+C< \pi \rightarrow k=0\rightarrow A-B=-C$ $P=2(\frac{1}{1+tan^2A}-\frac{1}{1+tan^2B})-\frac{4tanC}{\sqrt{1+tan^2C}}+\frac{3tanC}{(1+tan^2C)\sqrt{1+tan^2C}}$ $=2(cos^2A-cos^2B)-4sinC+3sin\times cos^2C$ $=cos2A-cos2B-4sinC+3sin\times cos^2C$ $=-2sin(A+B)sin(A-B)-4sinC+3sinCcos^2C$ $=2sinCsin(A+B)-4sinC+3sinCcos^2C$ $\Rightarrow P\leq 2sinC-4sinC+3sinCcos^2C=sinC(3cos^2C-2)=sinC(1-3sin^2C)$ If $sinC> \frac{\sqrt{3}}{3}\Rightarrow P< 0.$ ,suppose that $sinC\leq \frac{\sqrt{3}}{3}$ By applying AM-GM,we have: $P\leq \sqrt{sin^2C(1-3sin^C)^2}=\frac{\sqrt{6sin^2C(1-3sin^2C)(1-3sin^2C)}}{\sqrt{6}}$ $=\frac{\sqrt{\frac{6sin^2C+1-3sin^2C+1-3sin^2C}{3}}}{\sqrt{6}}$ $\LARGE =\frac{2}{9}=\frac{m}{n}{\color{#3D99F6} \rightarrow m+n=11}$