Let $O$ be the circumcenter of $ABC$ , by the conditions we have $\measuredangle BAX=\measuredangle BOX$ and $\measuredangle CAY=\measuredangle COY$ and by $\measuredangle BAX = \measuredangle CAY$ we have that $BX=CY$ ( We can do this directly from Law of Sines $BX=2R \sin(BAX)=2R \sin(CAY)=CY$ ) Now , from triangle inequality we have $AY-MB<AY-(BX+MX)=AY-BX-MX<AC-MX=AC-MA<MC$ $\Longleftrightarrow AY<MB+MC$

XbenX wrote: Let $O$ be the circumcenter of $ABC$ , by the conditions we have $\measuredangle BAX=\measuredangle BOX$ and $\measuredangle CAY=\measuredangle COY$ and by $\measuredangle BAX = \measuredangle CAY$ we have that $BX=CY$ ( We can do this directly from Law of Sines $BX=2R \sin(BAX)=2R \sin(CAY)=CY$ ) Now , from triangle inequality we have $AY-MB<AY-(BX+MX)=AY-BX-MX<AC-MX=AC-MA<MC$ $\Longleftrightarrow AY<MB+MC$ I think the first inequality should be the other way around, because $MB<BX+MX \implies -BX-MX<-MB \implies AY-BX-MX<AY-MB$.

See Iranian Geometry Olympiad 2014 Problem J5.