\(\begin{array}{l}
1)\,\frac{b}{{\sqrt {a + b} - \sqrt {a - b} }} = \frac{{b\left( {\sqrt {a + b} + \sqrt {a - b} } \right)}}{{\left( {\sqrt {a + b} - \sqrt {a - b} } \right)\left( {\sqrt {a + b} + \sqrt {a - b} } \right)}}\\
= \frac{{b\left( {\sqrt {a + b} + \sqrt {a - b} } \right)}}{{2b}} = \frac{{\sqrt {a + b} + \sqrt {a - b} }}{2}\\
\frac{c}{{\sqrt {a + c} - \sqrt {a - c} }} = \frac{{\sqrt {a + c} + \sqrt {a - c} }}{2}\\
Ta\,co:\,\frac{b}{{\sqrt {a + b} - \sqrt {a - b} }} < \frac{c}{{\sqrt {a + c} - \sqrt {a - c} }}\\
\Leftrightarrow \sqrt {a + b} + \sqrt {a - b} < \sqrt {a + c} + \sqrt {a - c} \\
\Leftrightarrow {\left( {\sqrt {a + b} + \sqrt {a - b} } \right)^2} < {\left( {\sqrt {a + c} + \sqrt {a - c} } \right)^2}\\
\Leftrightarrow 2a + 2\sqrt {{a^2} - {b^2}} < 2a + 2\sqrt {{a^2} - {c^2}} \\
\Leftrightarrow \sqrt {{a^2} - {b^2}} < \sqrt {{a^2} - {c^2}} \\
\Leftrightarrow b > c\,\left( {luôn\, đúng} \right)\\
\Rightarrow dpcm
\end{array}\)
