Question

[Nguyễn Hoàng Tiến]

Cho tam giác ABC , O nằm trong tam giác đó. Các tia AO,BO,CO cắt BC,CA,AB tại M,N,P. Chứng minh rằng:

\(\sqrt{\frac{OA}{OM}}+\sqrt{\frac{OB}{ON}}+\sqrt{\frac{OC}{OP}}\ge3\sqrt{2}\)

\(\sqrt{\frac{AM}{OA}}+\sqrt{\frac{BN}{OB}}+\sqrt{\frac{CP}{OC}}\ge\frac{3\sqrt{6}}{2}\)

\(\sqrt{\frac{OM}{AM}}+\sqrt{\frac{ON}{BN}}+\sqrt{\frac{OP}{CP}}\ge\sqrt{3}\)

Đã chứng minh:

\(\frac{AM}{OM}+\frac{BN}{ON}+\frac{CP}{OP}\ge9\)

\(\frac{OA}{AM}+\frac{OB}{ON}+\frac{OC}{OP}\ge6\)

\(\frac{AM}{OA}+\frac{BN}{OB}+\frac{CP}{OC}\ge\frac{9}{2}\)

\(\frac{OM}{OA}+\frac{ON}{OB}+\frac{OP}{OC}\ge\frac{3}{2}\)

 ( bài toán cực trị trong hình học). 

Answer 1

[Cô Hoàng Huyền]

a. Đặt \(S_{AOB}=c^2;S_{BOC}=a^2;S_{COA}=b^2\Rightarrow S_{ABC}=a^2+b^2+c^2\)

Ta có \(\frac{AM}{OM}=\frac{S_{ABC}}{S_{BOC}}=\frac{a^2+b^2+c^2}{a^2}=1+\frac{b^2+c^2}{a^2}\)

Vậy thì \(\frac{OA}{OM}=\frac{AM}{OM}-1=\frac{b^2+c^2}{a^2}\Rightarrow\sqrt{\frac{OA}{OM}}=\sqrt{\frac{b^2+c^2}{a^2}}\ge\frac{1}{\sqrt{2}}\left(\frac{b}{a}+\frac{a}{b}\right)\)

Tương tự, ta có: \(\sqrt{\frac{OA}{OM}}+\sqrt{\frac{OB}{ON}}+\sqrt{\frac{OC}{OP}}\ge\frac{1}{\sqrt{2}}\left(\frac{a}{b}+\frac{c}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}\right)\ge\frac{1}{\sqrt{2}}.6=3\sqrt{2}\)