It was indeed hard problem. But my solution is not unnatural for me. Here is my idea: I fix \angle{CBM} and find out when does \sin{\angle{CAM}} approach its maximum. Let's define: \angle{CBM}=\varphi and \angle{CAM}=\omega. First let's discuss the case when \varphi<\frac{\pi}{2}. We can also take BC=2 whithout loss of generality. Denote by K the midpoint of BC. Let O be a point lying on the ray CB so that OB=BC. Let l be a line passing through O and parallel to BM. All triangles that have \angle{CBM}=\varphi are formed by taking any point A on the line l on the same side as M is. Now we must determine the point A on line l such that \sin{\omega} is maximum. Now we take Decartese rectangle coordinate system (sorry if it is not the right word) with axis OX lying on line l such that points B,C have positive coordinates. Axis OY is perpendicular to OX with positive direction such that points B,C have positive coordinates. In this system we have B=(2,2\tan{\varphi}),K=(3,3\tan{\varphi}),C=(4,4\tan{\varphi}). The piont A which is moving by the positive side of OX has such coordinates A=(t,0) wheret>0. Let's calculate \vec{AK}\cdot\vec{AC}=(3-t)(4-t)+12{\tan{\varphi}}^{2}=t^{2}-7t+\frac{12}{{\cos{\varphi}}^{2}}. If \sin{\varphi}\leq\frac{1}{7} then there is such positive value of t when \vec{AK}\cdot\vec{AC}=0 i.e there is such point A such that \angle{CAM}=\frac{\pi}{2}, thus the maximum of \sin{\omega} is 1 so {\sin{\varphi}+\sin{\omega}}\leq{1+\frac{1}{7}}=\frac{8}{7}<\frac{2}{\sqrt[2]{3}}. If \sin{\varphi}>\frac{1}{7} then \vec{AK}\cdot\vec{AC}>0 i.e. \angle{CAM}<\frac{\pi}{2} so \sin{\omega} approaches its maximum when \tan{\omega} is maximum. \tan{\omega}=\tan(\angle{MAX}-\angle{CAX})=\frac{t\tan{\varphi}}{t^2-7t+\frac{12}{{\cos{\varphi}}^2}} so \tan{\omega} is maximum when t=\frac{12}{t{\cos{\varphi}}^{2}} and \tan{{\omega}_{max}}=\frac{\sin{\varphi}}{4\sqrt{3}-7\cos{\varphi}} and \sin{{\omega}_{max}}=\frac{\sin{\varphi}}{7-4\sqrt{3}\cos{\vaphi}}. So it we need to prove that \sin{\varphi}+{\sin{\varphi}}{7-4\sqrt{3}\cos{\varphi}}\leq\frac{2}{\sqrt{3}}. The last one is equvalent to {(\sin{\varphi}+\cos{\varphi}-\frac{2}{\sqrt{3}})}^{2}\geq{0} which is obvious and equality holds if only if \sin{\varphi}+\cos{\varphi}=\frac{2}{\sqrt{3}}. We are on. . Cases {\frac{\pi}{2}}\leq{\varphi}<{\pi} are done in the same way.

Oh, thank you, Levon Nurbekian, your solution is excellent (and it is also absolutely natural:)), it was such a pleasure for me to read it. In official solution they first consider the case when AB touches the circle AMC (in that case it is much easier to prove the inequality), and after that they go on to prove the general case using already proved case. I couldn't find any sensible explanation for such a structure of the proof.

Centroid $G$ of $\triangle ABC\ ,$ area $S=[ABC]\ ,\ \phi =m(\widehat {AGB})\ .$ Prove that $\sin\widehat {CAG}+\sin\widehat {CBG}\leq\frac{2}{\sqrt3}\Longleftrightarrow \boxed{\frac{a}{m_a}+\frac{b}{m_b}\le \frac{2ab}{S\sqrt 3}}\Longleftrightarrow \frac{m_a}{a}+\frac{m_b}{b}\le \frac{\sqrt 3}{\sin \phi}\ .$ An angular inequality is equivalently with a linear (more accessible) inequality ! See http://www.mathlinks.ro/Forum/viewtopic.php?t=83790