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.

$$\begin{cases}\sqrt{3-x}-\sqrt{x-1}>\sqrt{4-x}-\sqrt x\\ 1\le x\le 3\end{cases}$$
If you want to square both current sides, then you must first evaluate their signs, namely using $A>B\iff \begin{cases}A\ge 0\\ B\ge 0\ \\ A^2>B^2\end{cases}\lor \begin{cases}A\ge 0\\ B<0\end{cases}\lor\begin{cases}A< 0\\ B< 0\ \\ A^2<B^2\end{cases}$. I think that delaying such considerations by making both sides non-negative is better (recall that, in the reals, $\sqrt A\ge 0$ when defined):
\begin{align}&\begin{cases}\sqrt{3-x}+\sqrt x>\sqrt{4-x}+\sqrt{x-1}\\ 1\le x\le 3\end{cases}\\&\begin{cases}3+2\sqrt{x(3-x)}>3+2\sqrt{(4-x)(x-1)}\\ 1\le x\le 3\end{cases}\\ &\begin{cases}\sqrt{x(3-x)}>\sqrt{(4-x)(x-1)}\\ 1\le x\le 3\end{cases}\\&\begin{cases}x(3-x)>(4-x)(x-1)\\ 1\le x\le 3\end{cases}\end{align}
