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? 

One approach:
\begin{align}
\sqrt{ x + 2} - \sqrt{ x + 3} &< \sqrt{ 2} - \sqrt{ 3} \\
\sqrt{ x + 2}  + \sqrt{ 3} &<  \sqrt{ x + 3} + \sqrt{ 2}  \tag{1}\\
(x + 2)  + 3 + 2\sqrt{ 3}\sqrt{ x + 2}  &<  (x + 3) +  2 +2\sqrt{ 2}\sqrt{ x + 3} \tag{2}\\
2\sqrt{ 3}\sqrt{ x + 2}  &<  2\sqrt{ 2}\sqrt{ x + 3} \tag{3}\\
3(x + 2)  &<  2( x + 3) \tag{4}\\
x  &< 0 \tag{5}
\end{align}
where we

$(1)$ added $\sqrt{ x + 3} + \sqrt{3}$
$(2)$ squared both sides (they are positive)
$(3)$ subtracted $x+5$
$(4)$ divided by $2$ and squared (again both sides are positive)
$(5)$ subtracted $2x+6$

Now just combine it with the condition $x\geq -2$ and you are done.
