Cho tam giác ABC, M là điểm bất kì nằm trong tam g...

- Hướng dẫn giải

Phương pháp giải:

 Chứng minh \(\frac{MA}{MP}.\frac{MB}{MQ}.\frac{MC}{MK}\ge 8\)

Sử dụng tỉ lệ diện tích tam giác và bất đẳng thức Cô-si. 

Giải chi tiết:

Ta chứng minh \(\frac{MA}{MP}.\frac{MB}{MQ}.\frac{MC}{MK}\ge 8\). Đặt \({{S}_{1}}={{S}_{MBC}};\,\,{{S}_{2}}={{S}_{MAC}};\,\,{{S}_{3}}={{S}_{MAB}}\).

Ta có \(\frac{MA}{MP}=\frac{{{S}_{AMB}}}{{{S}_{PMB}}}=\frac{{{S}_{AMC}}}{{{S}_{PMC}}}=\frac{{{S}_{AMB}}+{{S}_{AMC}}}{{{S}_{PMB}}+{{S}_{PMC}}}=\frac{{{S}_{MAB}}+{{S}_{MAC}}}{{{S}_{MBC}}}=\frac{{{S}_{2}}+{{S}_{3}}}{{{S}_{1}}}\)

Tương tự ta chứng minh được \(\frac{MB}{MQ}=\frac{{{S}_{1}}+{{S}_{3}}}{{{S}_{2}}};\,\,\frac{MC}{MK}=\frac{{{S}_{1}}+{{S}_{2}}}{{{S}_{3}}}\) \(\begin{align} & \Rightarrow \frac{MA}{MP}.\frac{MB}{MQ}.\frac{MC}{MK}=\frac{{{S}_{2}}+{{S}_{3}}}{{{S}_{1}}}.\frac{{{S}_{1}}+{{S}_{3}}}{{{S}_{2}}}.\frac{{{S}_{1}}+{{S}_{2}}}{{{S}_{3}}} \\ & \Rightarrow {{\left( \frac{MA}{MP}.\frac{MB}{MQ}.\frac{MC}{MK} \right)}^{2}}={{\left( \frac{{{S}_{2}}+{{S}_{3}}}{{{S}_{1}}}.\frac{{{S}_{1}}+{{S}_{3}}}{{{S}_{2}}}.\frac{{{S}_{1}}+{{S}_{2}}}{{{S}_{3}}} \right)}^{2}}=\frac{{{S}_{2}}+{{S}_{3}}}{{{S}_{1}}}.\frac{{{S}_{1}}+{{S}_{3}}}{{{S}_{2}}}.\frac{{{S}_{1}}+{{S}_{2}}}{{{S}_{3}}}.\frac{{{S}_{2}}+{{S}_{3}}}{{{S}_{1}}}.\frac{{{S}_{1}}+{{S}_{3}}}{{{S}_{2}}}.\frac{{{S}_{1}}+{{S}_{2}}}{{{S}_{3}}} \\ & \Rightarrow {{\left( \frac{MA}{MP}.\frac{MB}{MQ}.\frac{MC}{MK} \right)}^{2}}=\frac{{{\left( {{S}_{1}}+{{S}_{2}} \right)}^{2}}}{{{S}_{1}}{{S}_{2}}}.\frac{{{\left( {{S}_{2}}+{{S}_{3}} \right)}^{2}}}{{{S}_{2}}{{S}_{3}}}.\frac{{{\left( {{S}_{1}}+{{S}_{3}} \right)}^{2}}}{{{S}_{1}}{{S}_{3}}}\overset{Cauchy}{\mathop{\ge }}\,\frac{{{\left( 2\sqrt{{{S}_{1}}{{S}_{2}}} \right)}^{2}}}{{{S}_{1}}{{S}_{2}}}.\frac{{{\left( 2\sqrt{{{S}_{2}}{{S}_{3}}} \right)}^{2}}}{{{S}_{2}}{{S}_{3}}}.\frac{{{\left( 2\sqrt{{{S}_{1}}{{S}_{3}}} \right)}^{2}}}{{{S}_{1}}{{S}_{3}}}={{4}^{3}}=64 \\ & \Rightarrow \frac{MA}{MP}.\frac{MB}{MQ}.\frac{MC}{MK}\ge 8 \\ & \Rightarrow MA.MB.MC\ge 8MP.MQ.MK\,\,\left( dpcm \right) \\ \end{align}\)