Let $F$ be a point inside a convex pentagon $ABCDE$, and let $a_{1}$, $a_{2}$, $a_{3}$, $a_{4}$, $a_{5}$ denote the distances from $F$ to the lines $AB$, $BC$, $CD$, $DE$, $EA$, respectively. The points $F_{1}$, $F_{2}$, $F_{3}$, $F_{4}$, $F_{5}$ are chosen on the inner bisectors of the angles $A$, $B$, $C$, $D$, $E$ of the pentagon respectively, so that $AF_{1} = AF$ , $BF_{2} = BF$ , $CF_{3} = CF$ , $DF_{4} = DF$ and $EF_{5} = EF$ . If the distances from $F_{1}$, $F_{2}$, $F_{3}$, $F_{4}$, $F_{5}$ to the lines $EA$, $AB$, $BC$, $CD$, $DE$ are $b_{1}$, $b_{2}$, $b_{3}$, $b_{4}$, $b_{5}$, respectively. Prove that $a_{1} + a_{2} + a_{3} + a_{4} + a_{5} \leq b_{1} + b_{2} + b_{3} + b_{4} + b_{5}$