We have, $\sqrt 13 > \sqrt 7$ & $\sqrt 12 > \sqrt 6$

$\therefore \sqrt 13 + \sqrt 12 > \sqrt 7 + \sqrt 6 $

$\Rightarrow \frac {1}{\sqrt 13 + \sqrt 12} < \frac {1}{\sqrt 7 + \sqrt 6}$

$\Rightarrow \frac {\sqrt 13 – \sqrt 12}{(\sqrt 13 + \sqrt 12)(\sqrt 13 – \sqrt 12)} < \frac {\sqrt 7 – \sqrt 6}{(\sqrt 7 + \sqrt 6)(\sqrt 7 – \sqrt 6)}$

$\therefore \sqrt 13 – \sqrt 12 < \sqrt 7 – \sqrt 6 $

LHS $ = \sqrt {25 + 3 + 2 \times 5 \times \sqrt 3 } – \sqrt {3 + 1 + 2 \times \sqrt 3 \times 1} $

$ = \sqrt {{5^2} + {{\sqrt 3 }^2} + 2 \times 5 \times \sqrt 3 } – \sqrt {{{\sqrt 3 }^2} + {1^2} + 2 \times \sqrt 3 \times 1} $

$ = \sqrt {{{(5 + \sqrt 3 )}^2}} – \sqrt {{{(\sqrt 3 + 1)}^2}} $

$ = (5 + \sqrt 3 ) – (\sqrt 3 + 1) = 4$

RHS $ = \sqrt 3 + \sqrt 2 \approx 3.14 \approx \pi$

Clearly, $4 > 3.14$