Solving $ \sqrt{3-x} - \sqrt{x-1} > \sqrt{4-x} - \sqrt{x} $ I am reviewing my high-school math (year 10-ish) but I am hitting a wall with this nasty inequality:
$$
\sqrt{3-x} - \sqrt{x-1} > \sqrt{4-x} - \sqrt{x}
$$
I find (steps below) the solution to be
$$
1 \le x \le 3 \text{ with } x \ne 2
$$
but the text-book and WolframAlpha both agree that the solution is
$$
1 \le x < 2
$$
I would really appreciate if you could point out my mistake.

I also tried using the patterns that the text-book offers in the theory section, $$\sqrt{A(x)}<B(x)$$ and $$\sqrt{A(x)}>B(x)$$ but the terms increased in number instead of decreasing. I can provide that derivation, but I suspect it goes in the wrong direction.

First of all for real cases, $$3\ge x\ge1$$
Observe that $x+3-x=x-1+4-x$
We need $$\sqrt x-\sqrt{x-1}>\sqrt{4-x}-\sqrt{3-x}$$
$$\iff\sqrt x+\sqrt{3-x}>\sqrt{4-x}+\sqrt{x-1}$$
As both sides are $>0$
we can safely take square in both sides
$$x+3-x+2\sqrt{x(3-x)}>4-x+x-1+2\sqrt{(4-x)(x-1)}$$
Again as both sides are $>0$
we can safely take square in both sides
$$x(3-x)>(4-x)(x-1)\iff 3x-x^2>4x-x^2-4+x\iff 2>x$$
