Cho đường thẳng $xx’$ và hai điểm $A, B$ nằm về một phía của nó, $M \in xx'$. Biết rằng :
$\frac{{\cos \widehat {AMx}}}{{\cos \widehat {BMx}}} = \frac{b}{a}\left( {a,b > 0} \right)$.
Chứng minh rằng với mọi điểm $N$ trên $xx’$ ta có:  $aNA + bNB \ge aMA + bMB$

\({\rm{aNA}} + bNB = \frac{a}{{AM}}NA.MA + \frac{b}{{BM}}NB.MB\)
$\begin{array}{l}
 \ge \frac{a}{{MA}}\overrightarrow {NA} .\overrightarrow {MA}  + \frac{b}{{MB}}\overrightarrow {NB} .\overrightarrow {MB} \\
 = \frac{{a\left( {\overrightarrow {NM}  + \overrightarrow {MA} } \right)\overrightarrow {MA} }}{{MA}} + \frac{{b\left( {\overrightarrow {NM}  + \overrightarrow {MB} } \right)\overrightarrow {MB} }}{{MB}}\\
 = \overrightarrow {NM} \left( {a\frac{{\overrightarrow {MA} }}{{MA}} + b\frac{{\overrightarrow {MB} }}{{MB}}} \right) + {\rm{a}}{\rm{.MA}} + b.MB\\
 = a|\overrightarrow {NM} ||\frac{{\overrightarrow {MA} }}{{MA}}|.c{\rm{os}}\left( {\overrightarrow {NM} ,\frac{{\overrightarrow {MA} }}{{MA}}} \right) + b|\overrightarrow {NM} |.|\frac{{\overrightarrow {MB} }}{{MB}}|.c{\rm{os}}\left( {\overrightarrow {NM} ,\frac{{\overrightarrow {MB} }}{{MB}}} \right) + {\rm{aMA}} + bMB\\
 = |\overrightarrow {NM} |\left( {a\cos \left( {\overrightarrow {NM} ,\overrightarrow {MA} } \right) + b\cos \left( {\overrightarrow {NM} ,\overrightarrow {MB} } \right)} \right) + aMA + bMB
\end{array}$
Nếu $N \in Mx$ thì
$a\cos \left( {\overrightarrow {NM} ,\overrightarrow {MA} } \right) + b\cos \left( {\overrightarrow {NM} ,\overrightarrow {MB} } \right) =  - a\cos \widehat {{\rm{AMx}}} + b\cos \widehat {BMx'} = 0$
Nếu $N \in Mx'$, tương tự ta có
$a\cos \left( {\overrightarrow {NM} ,\overrightarrow {MA} } \right) + b\cos \left( {\overrightarrow {NM} ,\overrightarrow {MB} } \right) = 0$
Vậy \({\rm{aNA}} + bNB \ge {\rm{aMA}} + bMB\)
Đẳng thức xảy ra $ \Leftrightarrow \left\{ \begin{array}{l}
\overrightarrow {NA}  \text {cùng hướng với}  \overrightarrow {MA} \\
\overrightarrow {NB}   \text {cùng hướng với}  \overrightarrow {MB}
\end{array} \right. \Rightarrow N \equiv M$

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