Đáp án: $\min=0$ khi $\Delta{ABC}$ là tam giác đều
Giải thích các bước giải:
Vì $BC$ cố định nên ta chỉ việc tìm GTNN của $P=\dfrac{2}{A{{B}^{2}}}+\dfrac{2}{A{{C}^{2}}}-\dfrac{3}{A{{D}^{2}}}$
Kẻ $BH\bot AC$$\Rightarrow \cos A=\dfrac{AH}{AB}\Rightarrow AH=AB.\cos A$
Định lý Pitago, ta có:$\begin{cases}BH^2=BC^2-CH^2\\BH^2=AB^2-AH^2\end{cases}$
$\Leftrightarrow B{{C}^{2}}-C{{H}^{2}}=A{{B}^{2}}-A{{H}^{2}}$
$\Leftrightarrow B{{C}^{2}}=A{{B}^{2}}-A{{H}^{2}}+{{\left( AC-AH \right)}^{2}}$
$\Leftrightarrow B{{C}^{2}}=A{{B}^{2}}-A{{H}^{2}}+\left( A{{C}^{2}}-2AC.AH+A{{H}^{2}} \right)$
$\Leftrightarrow B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}-2AC.AH$
$\Leftrightarrow B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}-2AB.AC.\cos A$
$\Leftrightarrow B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}-2AB.AC.\cos 60{}^\circ $
$\Leftrightarrow B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}-AB.AC$
………………………………….
Có: $\dfrac{BD}{AB}=\dfrac{CD}{AC}=\dfrac{BD+CD}{AB+AC}=\dfrac{BC}{AB+AC}$
$\Rightarrow \dfrac{BD}{AB}\cdot \dfrac{CD}{AC}={{\left( \dfrac{BC}{AB+AC} \right)}^{2}}$
$\Rightarrow BD.CD=\dfrac{AB.AC.B{{C}^{2}}}{{{\left( AB+AC \right)}^{2}}}$
………………………………….
Lấy điểm $E\in AC$ sao cho $\widehat{ABD}=\widehat{ADE}$
Khi đó, có hai cặp tam giác đồng dạng
$\Delta ABD\backsim\Delta ADE$ và $\Delta CED\backsim\Delta CDA$
$\Rightarrow \dfrac{AB}{AD}=\dfrac{AD}{AE}$ và $\dfrac{CE}{CD}=\dfrac{CD}{AC}=\dfrac{BD}{AB}$
$\Rightarrow A{{D}^{2}}=AB.AE$ và $BD.CD=AB.CE$
$\Rightarrow A{{D}^{2}}=AB\left( AC-CE \right)$
$\Rightarrow A{{D}^{2}}=AB.AC-AB.CE$
$\Rightarrow A{{D}^{2}}=AB.AC-BD.CD$
$\Rightarrow A{{D}^{2}}=AB.AC-\dfrac{AB.AC.B{{C}^{2}}}{{{\left( AB+AC \right)}^{2}}}$
$\Rightarrow A{{D}^{2}}=AB.AC\left[ 1-\dfrac{B{{C}^{2}}}{{{\left( AB+AC \right)}^{2}}} \right]$
$\Rightarrow A{{D}^{2}}=AB.AC\cdot \dfrac{{{\left( AB+AC \right)}^{2}}-B{{C}^{2}}}{{{\left( AB+AC \right)}^{2}}}$
$\Rightarrow A{{D}^{2}}=AB.AC.\dfrac{{{\left( AB+AC \right)}^{2}}-\left( A{{B}^{2}}+A{{C}^{2}}-AB.AC \right)}{{{\left( AB+AC \right)}^{2}}}$
$\Rightarrow A{{D}^{2}}=AB.AC\cdot \dfrac{3AB.AC}{{{\left( AB+AC \right)}^{2}}}$
$\Rightarrow A{{D}^{2}}=\dfrac{3A{{B}^{2}}.A{{C}^{2}}}{{{\left( AB+AC \right)}^{2}}}$
$\Rightarrow \dfrac{3}{A{{D}^{2}}}=\dfrac{{{\left( AB+AC \right)}^{2}}}{A{{B}^{2}}.A{{C}^{2}}}$
………………………………….
$P=\dfrac{2}{A{{B}^{2}}}+\dfrac{2}{A{{C}^{2}}}-\dfrac{3}{A{{D}^{2}}}$
$P=\dfrac{2}{A{{B}^{2}}}+\dfrac{2}{A{{C}^{2}}}-\dfrac{{{\left( AB+AC \right)}^{2}}}{A{{B}^{2}}.A{{C}^{2}}}$
$P=\dfrac{2A{{C}^{2}}+2A{{B}^{2}}-{{\left( AB+AC \right)}^{2}}}{A{{B}^{2}}.A{{C}^{2}}}$
$P=\dfrac{{{\left( AB-AC \right)}^{2}}}{A{{B}^{2}}.A{{C}^{2}}}\ge 0$
$\Rightarrow \min =0$
Dấu “=” xảy ra khi $AB=AC\Leftrightarrow \Delta ABC$ là tam giác đều
