Bonsoir,
Hypothèse:
0<a<b<c
2b=a+c
Thèse:
[tex] \dfrac{2}{ \sqrt{a}+ \sqrt{c} } = \dfrac{1}{ \sqrt{a}+ \sqrt{b}} + \dfrac{1}{ \sqrt{b}+ \sqrt{c}} [/tex]
Démonstration:
[tex]2b=a+c\\\\
2b+ 2\sqrt{ab} +2 \sqrt{ac} +2 \sqrt{bc} =a+c+ \sqrt{ac} +2 \sqrt{ab}+ \sqrt{ac} +2 \sqrt{bc} \\\\
2( \sqrt{a}* \sqrt{b} +b+ \sqrt{a}* \sqrt{c} + \sqrt{b}* \sqrt{c} )=\\
(\sqrt{a}* \sqrt{a} + \sqrt{a} * \sqrt{c} +2 \sqrt{a} * \sqrt{b})+ (\sqrt{c}* \sqrt{a} + \sqrt{c} * \sqrt{c}+2 \sqrt{c} * \sqrt{b} )\\\\
[/tex]
[tex]2( \sqrt{a}+ \sqrt{b})( \sqrt{b}+ \sqrt{c} )=\\
\sqrt{a}*( \sqrt{b} + \sqrt{c} + \sqrt{a} + \sqrt{b})+ \sqrt{c}*(\sqrt{b}+ \sqrt{c} + \sqrt{a} + \sqrt{b} )\\\\
2( \sqrt{a}+ \sqrt{b})( \sqrt{b}+ \sqrt{c} )=(\sqrt{a}+\sqrt{c}) *( \sqrt{b} + \sqrt{c} + \sqrt{a} + \sqrt{b})\\\\
\dfrac{2}{\sqrt{a} + \sqrt{c} } = \dfrac{( \sqrt{b} + \sqrt{c} + \sqrt{a} + \sqrt{b})}{(\sqrt{a}+ \sqrt{b})( \sqrt{b}+ \sqrt{c} ) } \\\\
[/tex]
[tex]\dfrac{2}{\sqrt{a} + \sqrt{c} } =\dfrac{1 }{ \sqrt{a}+ \sqrt{b}}+\dfrac{1 }{\sqrt{b}+ \sqrt{c}}\\\\
[/tex]
