Prove that,$\sqrt {28 + 10\sqrt 3 } - \sqrt {4 + 2\sqrt 3 } $$> \sqrt 3 + \sqrt 2 $ or $> \pi$

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, $4 > 3.14$