<s>[INST] 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? 
 [/INST] First lets multiply the inequality by $-1$ so we can square. $$\sqrt{x+2}-\sqrt{x+3}<\sqrt2-\sqrt3\\\sqrt{x+3}-\sqrt{x+2}>\sqrt{3}-\sqrt{2}$$Now lets square$$2x+5-2\sqrt{(x+3)(x+2)}>5-2\sqrt{6}\\x+\sqrt6>\sqrt{(x+3)(x+2)}$$
Now since $x+\sqrt{6}\geq-2+\sqrt 6>0$ we can square the inequality again
$$x^2+2\sqrt{6}x+6>x^2+5x+6\\(2\sqrt6-5)x>0$$
This happens when $x<0$ since $2\sqrt6=\sqrt{24}<\sqrt{25}=5$ is negative so $x$ must be as well.
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