a > b > 0 {\displaystyle a>b>0} としたとき、 a + b + 2 a b = a + b {\displaystyle {\sqrt {a+b+2{\sqrt {ab}}}}={\sqrt {a}}+{\sqrt {b}}} , a + b − 2 a b = a − b {\displaystyle {\sqrt {a+b-2{\sqrt {ab}}}}={\sqrt {a}}-{\sqrt {b}}}
となることを証明する。
a + b + 2 a b = ( a ) 2 + 2 a b + ( b ) 2 = ( a + b ) 2 = a + b {\displaystyle {\begin{aligned}{\sqrt {a+b+2{\sqrt {ab}}}}&={\sqrt {({\sqrt {a}})^{2}+2{\sqrt {a}}{\sqrt {b}}+({\sqrt {b}})^{2}}}\\&={\sqrt {({\sqrt {a}}+{\sqrt {b}})^{2}}}\\&={\sqrt {a}}+{\sqrt {b}}\end{aligned}}} , a + b − 2 a b = ( a ) 2 − 2 a b + ( b ) 2 = ( a − b ) 2 = a − b {\displaystyle {\begin{aligned}{\sqrt {a+b-2{\sqrt {ab}}}}&={\sqrt {({\sqrt {a}})^{2}-2{\sqrt {a}}{\sqrt {b}}+({\sqrt {b}})^{2}}}\\&={\sqrt {({\sqrt {a}}-{\sqrt {b}})^{2}}}\\&={\sqrt {a}}-{\sqrt {b}}\end{aligned}}}
