SAMC

Let $K$ be a point on the ray $BN$ such that $\widehat{BMA}=\widehat{BCK}$. It is clear that $K$ is outside of triangle because $\widehat{BMA}>\widehat{ACB}$. We have $△ABM\sim △KBC$, hence $$\frac{AB}{BK}=\frac{BM}{BC}=\frac{AM}{CK}.\text{\hspace{1em}(1)}$$ Now, we have $△ABK\sim △MBC$ (since $\widehat{ABK}=\widehat{MBC}$ and $\frac{AB}{BK}=\frac{BM}{BC}$), hence $$\frac{AB}{BM}=\frac{BK}{BC}=\frac{AK}{CM}.\text{\hspace{1em}(2)}$$ We obtained $\widehat{CKN}=\widehat{MAB}=\widehat{NAC}$. It follows that quadrilateral $ANCK$ is cyclic, and by using Ptolemy's theorem we get $$AC\cdot (BK-BN)=AN\cdot CK+CN\cdot AK.\text{\hspace{1em}(3)}$$ From (1) and (2) it follows $$CK=\frac{AM\cdot BC}{BM},AK=\frac{AB\cdot CM}{BM},BK=\frac{AB\cdot BC}{BM}.$$ Replacing these relations in (3) we get $$AC\cdot \left(\frac{AB\cdot BC}{BM}-BN\right)=\frac{AM\cdot AN\cdot BC}{BM}+\frac{CM\cdot CN\cdot AB}{BM},$$ hence \frac{AM \cdot AN}{AB \cdot AC} + \frac{BM \cdot BN}{BA \cdot BC} + \frac{CM \cdot CN}{CA \cdot CB} = 1 . --- **Alternative solution.** Using Ceva's theorem in trigonometric form for points $M$ and $N$ we get \frac{\sin \widehat{MAB}}{\sin \widehat{MAC}} \cdot \frac{\sin \widehat{MBC}}{\sin \widehat{MBA}} \cdot \frac{\sin \widehat{MCA}}{\sin \widehat{MCB}} = 1, $$and$$ \frac{\sin \widehat{NAB}}{\sin \widehat{NAC}} \cdot \frac{\sin \widehat{NBC}}{\sin \widehat{NBA}} \cdot \frac{\sin \widehat{NCA}}{\sin \widehat{NCB}} = 1 $$Multiplyingtheserelations,itfollows$$ \frac{\sin \widehat{MCA}}{\sin \widehat{MBC}} \cdot \frac{\sin \widehat{NCA}}{\sin \widehat{NCB}} = 1, $$hence$$ \frac{\sin \widehat{MCA}}{\sin \widehat{MCB}} = \frac{\sin \widehat{NCB}}{\sin \widehat{NCA}} . \tag{2} Denote $\widehat{MCB} = \alpha$ and $\widehat{MCA} = \beta$ and get \frac{\sin (C - \alpha)}{\sin \alpha} = \frac{\sin (C - \beta)}{\sin \beta} . \tag{3} $$or$$ \sin C \cdot \cot \alpha - \cos C = \sin C \cdot \cot \beta - \cos C hence $\alpha = \beta$. Using complex coordinates, relation $\widehat{MCB} = \widehat{ACN}$ shows that \arg \frac{m-c}{b-c} = \arg \frac{a-c}{n-c}, hence the number $\frac{m-c}{b-c} : \frac{a-c}{n-c}$ is real. We have \frac{m-c}{b-c} \cdot \frac{n-c}{a-c} = \left| \frac{m-c}{b-c} \cdot \frac{n-c}{a-c} \right| = \frac{CM \cdot CN}{CB \cdot CA} . $$Finally,weget$$ \sum \frac{AM \cdot AN}{AB \cdot AC} = \sum \frac{(m-a)(n-a)}{(b-a)(c-a)} \\ = \frac{\sum (m-a)(n-a)(b-c)}{(a-b)(b-c)(c-a)} = 1 $$