Question

Solve the following inequality:
$\sqrt{x+3}<\sqrt{x-1}+\sqrt{x-2}.$

Answer 1

$=>\sqrt{x+3}<\sqrt{x-1}+\sqrt{x-2}$

On squaring we have

$=>x+3<x-1+x-2+2\sqrt{(x-1)(x-2)}$

$=>0<2x-x-3-3+2\sqrt{(x-1)(x-2)}$

$=>0<x-6+2\sqrt{(x-1)(x-2)}$

$=>-2\sqrt{x^2-3x+2}<x-6$

$=>4(x^2-3x+2)<x^2+36-12x$

$=>{4x}^2-x^2+12x-12x+8-36<0$

$=>{3x}^2-28<0(x+\sqrt{\frac{28}{3})}(x-\sqrt{\frac{28}{3})}<0$

$=>xϵ(-\sqrt{\frac{28}{3}},\sqrt{\frac{28}{3}})$

but, $x>2$

Hence, $xϵ(2,\sqrt{\frac{28}{3}})$.