Question

Cho tam giác ABC. Chứng minh rằng

\(\dfrac{h_b}{h_a^2}+\dfrac{h_c}{h_b^2}+\dfrac{h_a}{h_c^2}>\dfrac{1}{r}\)

Answer 1

\(\dfrac{h_b}{h_a^2}+\dfrac{h_c}{h_b^2}+\dfrac{h_a}{h_c^2}=\dfrac{\dfrac{2S_{ABC}}{b}}{\dfrac{4S_{ABC}^2}{a^2}}+\dfrac{\dfrac{2S_{ABC}}{c}}{\dfrac{4S^2_{ABC}}{b^2}}+\dfrac{\dfrac{2S_{ABC}}{a}}{\dfrac{4S_{ABC}^2}{c^2}}\)

\(=\dfrac{a^2}{2bS_{ABC}}+\dfrac{b^2}{2cS_{ABC}}+\dfrac{c^2}{2aS_{ABC}}\)

\(=\dfrac{1}{2S_{ABC}}\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\right)\)

\(\ge\dfrac{1}{2.\dfrac{a+b+c}{2}r}.\dfrac{\left(a+b+c\right)^2}{a+b+c}=\dfrac{1}{r}\)

Hình như có dấu = chứ nhỉ

Đẳng thức xảy ra khi tam giác ABC đều