<s>[INST] Determine all real $x$ that satisfies $\sqrt{3-x} - \sqrt{x+1} > \frac{1}{2}$ I have this equation that I have to solve:

Determine all real $x$ that satisfies $\sqrt{3-x} - \sqrt{x+1} > \frac{1}{2}$

Maybe it involves means? The square roots are leaning towards it... Could someone help me?
 [/INST] Note that


*

*$\sqrt{3-x}$ is defined only for $x\le3$

*$\sqrt{x+1}$ is defined only for $x\ge-1$

*The first square root is strictly decreasing and the second strictly increasing, so the whole LHS must be strictly decreasing (to see why, graph the functions)


Thus it remains to solve for $x$ in the case of equality, and take the interval as $[-1,x)$.
$$\sqrt{3-x}-\sqrt{x+1}=\frac12$$
$$\sqrt{3-x}=\frac12+\sqrt{x+1}$$
$$3-x=\frac14+\sqrt{x+1}+x+1$$
$$\frac74-2x=\sqrt{x+1}$$
$$\frac{49}{16}-7x+4x^2=x+1$$
$$4x^2-8x+\frac{33}{16}=0$$
$$x=1\pm\frac{\sqrt{31}}8$$
Checking, we see that the negative sign must be used for $x$, as only that produces an equality. Hence the range of $x$ is $[-1,1-\frac{\sqrt{31}}8)$.
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