Câu hỏi của Hỏi Làm Gì

Question

Cho tam giác ABC nhọn , O là điểm nằm trong tam giác. Các tia AO, BO, CO lần lượt cắt BC, AC, AB tại M,N,P. Chứng minh rằng:
$\frac{AM}{OM}+\frac{BN}{ON}+\frac{CP}{OP}\ge9$
Giúp với bà con!!!

Hoàng Lê Bảo Ngọc · Accepted Answer

A B C M N P O  Ta có : $\frac{OM}{AM}=\frac{S_{BOC}}{S_{ABC}}$ ; $\frac{ON}{BN}=\frac{S_{AOC}}{S_{ABC}}$ ; $\frac{OP}{CP}=\frac{S_{AOB}}{S_{ABC}}$$\Rightarrow\frac{OM}{AM}+\frac{ON}{BN}+\frac{OP}{CP}=\frac{S_{ABC}}{S_{ABC}}=1$Áp dụng bđt Bunhiacopxki, ta có : $\frac{AM}{OM}+\frac{BN}{ON}+\frac{CP}{OP}=\left(\frac{AM}{OM}+\frac{BN}{ON}+\frac{CP}{OP}\right).\left(\frac{OM}{AM}+\frac{ON}{BN}+\frac{OP}{CP}\right)\ge$$\ge\left(\sqrt{\frac{AM}{OM}.\frac{OM}{AM}}+\sqrt{\frac{BN}{ON}.\frac{ON}{BN}}+\sqrt{\frac{CP}{OP}.\frac{OP}{CP}}\right)^2=\left(1+1+1\right)^2=9$Vậy $\frac{AM}{OM}+\frac{BN}{ON}+\frac{CP}{OP}\ge9$ (đpcm)Câu trả lời: A B C M N P O  Ta có : $\frac{OM}{AM}=\frac{S_{BOC}}{S_{ABC}}$ ; $\frac{ON}{BN}=\frac{S_{AOC}}{S_{ABC}}$ ; $\frac{OP}{CP}=\frac{S_{AOB}}{S_{ABC}}$$\Rightarrow\frac{OM}{AM}+\frac{ON}{BN}+\frac{OP}{CP}=\frac{S_{ABC}}{S_{ABC}}=1$Áp dụng bđt Bunhiacopxki, ta có : $\frac{AM}{OM}+\frac{BN}{ON}+\frac{CP}{OP}=\left(\frac{AM}{OM}+\frac{BN}{ON}+\frac{CP}{OP}\right).\left(\frac{OM}{AM}+\frac{ON}{BN}+\frac{OP}{CP}\right)\ge$$\ge\left(\sqrt{\frac{AM}{OM}.\frac{OM}{AM}}+\sqrt{\frac{BN}{ON}.\frac{ON}{BN}}+\sqrt{\frac{CP}{OP}.\frac{OP}{CP}}\right)^2=\left(1+1+1\right)^2=9$Vậy $\frac{AM}{OM}+\frac{BN}{ON}+\frac{CP}{OP}\ge9$ (đpcm)