Đặt $BH=x\left( x>0 \right)$. Ta có

$BD=\sqrt{D{{H}^{2}}+B{{H}^{2}}}=\sqrt{{{x}^{2}}+16}$

Vì $DH//AC$ nên

$\frac{DA}{DB}=\frac{HC}{HB}\Rightarrow DA=\frac{DB.HC}{HB}=\frac{\sqrt{{{x}^{2}}+16}}{2x}$

$\Rightarrow AB=\sqrt{{{x}^{2}}+16}+\frac{\sqrt{{{x}^{2}}+16}}{2x}$

Xét hàm số  $f\left( x \right)=\sqrt{{{x}^{2}}+16}+\frac{\sqrt{{{x}^{2}}+16}}{2x}$ trên $\left( 0;+\infty  \right)$. Ta có f(x) liên tục trên $\left( 0;+\infty  \right)$ và

$f'\left( x \right)=\frac{x}{\sqrt{{{x}^{2}}+16}}+\frac{\frac{x}{\sqrt{{{x}^{2}}+16}}.2x-2\sqrt{{{x}^{2}}+16}}{4{{x}^{2}}}=\frac{x}{\sqrt{{{x}^{2}}+16}}-\frac{8}{{{x}^{2}}\sqrt{{{x}^{2}}+16}}=\frac{{{x}^{3}}-8}{{{x}^{2}}\sqrt{{{x}^{2}}+16}}$

$f'\left( x \right)=0\Leftrightarrow x=2;f'\left( x \right)>0\Leftrightarrow x>2;f'\left( x \right)<0\Leftrightarrow 0

Suy ra $\min \,AB=\underset{x\in \left( 0;+\infty  \right)}{\mathop{\min }}\,f\left( x \right)=f\left( 2 \right)=\frac{5\sqrt{5}}{2}\approx 5,5902\,\left( m \right)$