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? 
 A: 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 $x$.
A: 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.
A: 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.
