Add 2 and then multiply with BN/BM, we have $\frac{BN_1 \cdot BN}{BM_1 \cdot BM}+\frac{BN_2 \cdot BN}{BM_2 \cdot BM} = \frac{S(BNN_1)}{S(BMM_1)}+\frac{S(BNN_2)}{S(BMM_2)}$ Since $\displaystyle S(BMM_1)=S(BMM_2)=\frac12 S(BM_1M_2)$, so $\frac{S(BNN_1)}{S(BMM_1)}+\frac{S(BNN_2)}{S(BMM_2)} = \frac{2S(BN_1N_2)}{S(BM_1M_2)} = \frac{2BN_1 \cdot BN_2}{BM_1 \cdot BM_2}$ Thus we only need to show that $\frac{BN_1 \cdot BN_2}{BM_1 \cdot BM_2} \geq \frac{BN}{BM}$ (remember we added 2 to both sides) Let $M_1'$, $M_2'$ and $M'$ be the projections on BC. Since <B=90, so the inequality becomes $BM'^2 \geq BM_1' \cdot BM_2'$. Since M' is the midpoint of $M_1'M_2'$, so $2BM'=BM_1'+BM_2' \geq 2\sqrt{BM_1' \cdot BM_2'}$. Done