Let $MA'=d_{a},MB'=d_{b},MC'=d_{c},\angle MAB=\alpha_{c},\angle MAC=\alpha_{b}$ we have $MB\cdot MC=\frac{d_{b}d_{c}}{\sin\alpha_{c}\sin\alpha_{b}}=\frac{2d_{b}d_{c}}{\cos(\alpha_{c}-\alpha_{b})-\cos(\alpha_{c}+\alpha_{b})}\ge \frac{2d_{b}d_{c}}{1-\cos A}=\frac{d_{b}d_{c}}{\sin^{2}\frac{A}{2}}$ similarly we get $MC\cdot MA\ge \frac{d_{c}d_{a}}{\sin^{2}\frac{B}{2}},MA\cdot MB\ge \frac{d_{a}d_{b}}{\sin^{2}\frac{C}{2}}$ multiply them and square root them we get $\frac{d_{a}d_{b}d_{c}}{MA\cdot MB\cdot MC}\le \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$ Equality holds iff $M\equiv I$ incenter of $ABC$.

this is also in imo 01 shortlist... does the hungary-israel binational competition takes its problems from other sources?

let x = MÂC', y = MBA' and z = MCB' so, sin(x) = MA'/MA sin(A-x) = MB'/MA,... so (MA'.MB'.MC'/MA.MB.MC)²=sin(x)sin(A-x)sin(y)sin(B-y)sin(z)sin(C-z) or sin(x)sin(A-x)<=(sin(x)+sin(A-x)/2)² (AM-GM) <=(sin(A/2))² by Jensen so : MA'.MB'.MC'/MA.MB.MC <=sin(A/2)sin(B/2)sin(C/2) with equality if M=I