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\(\dfrac{h_b}{h_a^2}+\dfrac{h_c}{h_b^2}+\dfrac{h_a}{h_c^2}=\dfrac{\dfrac{2S_{ABC}}{b}}{\dfrac{4S_{ABC}^2}{a^2}}+\dfrac{\dfrac{2S_{ABC}}{c}}{\dfrac{4S^2_{ABC}}{b^2}}+\dfrac{\dfrac{2S_{ABC}}{a}}{\dfrac{4S_{ABC}^2}{c^2}}\)
\(=\dfrac{a^2}{2bS_{ABC}}+\dfrac{b^2}{2cS_{ABC}}+\dfrac{c^2}{2aS_{ABC}}\)
\(=\dfrac{1}{2S_{ABC}}\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\right)\)
\(\ge\dfrac{1}{2.\dfrac{a+b+c}{2}r}.\dfrac{\left(a+b+c\right)^2}{a+b+c}=\dfrac{1}{r}\)
Hình như có dấu = chứ nhỉ
Đẳng thức xảy ra khi tam giác ABC đều
Ta có : \(\dfrac{1}{2}\sqrt{\overrightarrow{AB}^2\overrightarrow{AC}^2-\left(\overrightarrow{AB}.\overrightarrow{AC}\right)^2}\)
\(=\dfrac{1}{2}.\sqrt{AB^2AC^2-\left(AB.AC.CosBAC\right)^2}\)
\(=\dfrac{1}{2}.\sqrt{AB^2AC^2-AB^2.AC^2.Cos^2BAC}\)
\(=\dfrac{1}{2}\sqrt{AB^2AC^2\left(1-Cos^2BAC\right)}\)
Thấy : \(Sin^2a+Cos^2a=1\)
\(\Rightarrow Sin^2a=1-Cos^2a\)
\(\Rightarrow\dfrac{1}{2}\sqrt{AB^2AC^2Sin^2BAC}=\dfrac{1}{2}\left|AB.AC.SinBAC\right|=\dfrac{1}{2}AB.AC.SinBAC=S\)
=> ĐPCM
Sao đề là lạ đoạn kia là \(\left(\overrightarrow{AB}.\overrightarrow{AC}\right)^2\)à
\(a=2b-2c\Rightarrow sinA.2R=2sinB.2R-2sinC.2R\)
\(\Rightarrow sinA=2sinB-2sinC\)
\(ah_a=bh_b=ch_c\Rightarrow\left(2b-2c\right)h_a=bh_b=ch_c\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{h_a}=\dfrac{2b-2c}{b}.\dfrac{1}{h_b}\\\dfrac{1}{h_a}=\dfrac{2b-2c}{c}.\dfrac{1}{h_c}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{h_a}=\dfrac{1}{h_b}-\dfrac{1}{h_c}+\left(\dfrac{b}{c.h_c}-\dfrac{c}{b.h_b}\right)\)
Câu này đề sai tiếp, biểu thức \(\dfrac{b}{c.h_c}-\dfrac{c}{b.h_b}\) kia không thể bằng 0
Đặt \(A = \dfrac{1}{2}\sqrt {{{\overrightarrow {AB} }^2}.{{\overrightarrow {AC} }^2} - {{\left( {\overrightarrow {AB} .\overrightarrow {AC} } \right)}^2}} \)
\(= \dfrac{1}{2}\sqrt { A{B^2}.A{C^2}- {{\left(|{\overrightarrow {AB}| .|\overrightarrow {AC}|. \cos BAC} \right)}^2}} \)
\(\begin{array}{l} \Rightarrow A = \dfrac{1}{2}\sqrt {A{B^2}.A{C^2} - {{\left( {AB.AC.\cos A} \right)}^2}} \\ \Leftrightarrow A = \dfrac{1}{2}\sqrt {A{B^2}.A{C^2} - A{B^2}.A{C^2}.{{\cos }^2}A }\\ \Leftrightarrow A = \dfrac{1}{2}\sqrt {A{B^2}.A{C^2}\left( {1 - {{\cos }^2}A} \right)} \end{array}\)
Mà \(1 - {\cos ^2}A = {\sin ^2}A\)
\( \Rightarrow A = \dfrac{1}{2}\sqrt {A{B^2}.A{C^2}.{{\sin }^2}A} \)
\( \Leftrightarrow A = \dfrac{1}{2}.AB.AC.\sin A\) (Vì \({0^o} < \widehat A < {180^o}\) nên \(\sin A > 0\))
Do đó \(A = {S_{ABC}}\) hay \({S_{ABC}} = \dfrac{1}{2}\sqrt {{{\overrightarrow {AB} }^2}.{{\overrightarrow {AC} }^2} - {{\left( {\overrightarrow {AB} .\overrightarrow {AC} } \right)}^2}} .\) (đpcm)