Solve the inequality $\sqrt { x + 2} - \sqrt { x + 3} < \sqrt { 2} - \sqrt { 3}$ Solve for $x$ real the inequality $$\sqrt { x + 2} - \sqrt { x + 3} < \sqrt { 2} - \sqrt { 3}.$$ Obviously $x\ge-2$. After that I tried to square the whole inequality, which led me to $x < - \frac { 18} { 4\sqrt { 6} - 5}$. Now, the answer is $[-2;0) $. Should there be a different approach? 

If we differentiate, $\frac d{dx}(\sqrt{x+2}-\sqrt{x+3})=\frac{1}{2}\Big(\frac1{\sqrt{x+2}}-\frac1{\sqrt{x+3}}\Big)>0$. So, provided $x\geq-2$, the LHS is continuous and increasing with $x$ and it will be equal to the RHS when $x=0$. Therefore it is less than the RHS when $-2\leq x<0$.
To see it is increasing without calculus, note that
$$\sqrt{x+2}-\sqrt{x+3}=\frac{-1}{\sqrt{x+2}+\sqrt{x+3}},$$
and the denominator is clearly increasing with $x$.
