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.
 A: Rearrange like $$
\sqrt{3-x}+ \sqrt{x} > \sqrt{x-1} + \sqrt{4-x}
$$
and now you can square it. It is easy know, since a lot of terms cancels out.
A: 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
$$x(3-x)>(4-x)(x-1)\iff 3x-x^2>4x-x^2-4+x\iff 2>x$$
A: $$\begin{cases}\sqrt{3-x}-\sqrt{x-1}>\sqrt{4-x}-\sqrt x\\ 1\le x\le 3\end{cases}$$
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}
A: You noticed the problem at $x=2$, where both sides become $=0$. This might have tipped you to not that for $x>2$, both sides become negative. Thereby squaring reverses the order relation in that range.
Alternatively, with differences of square roots, it is often helpful to use $\sqrt A-\sqrt B=\frac{A-B}{\sqrt A+\sqrt B}$ and thereby end up with sums of squarte roots instead. Here, this trick would turn the given inequality into
$$ \frac{4-2x}{\sqrt{3-x}+\sqrt {x-1}}>\frac{4-2x}{\sqrt{4-x}+\sqrt {x}}$$
This is not easier to solve, but if you had tried, you would once more be hinted to the sign change at $x=2$. 
At any rate, you would have been more careful and rearranged terms in a way that makes both sides positive (as in User2020201's answer) to allow squaring without harm.
A: The main problem is that you're squaring both sides of an inequality, and that is not allowed in general. You can do that only if the two side are positive:  $-5<1$ is true, but $25<1$ is false. So you have to add two condition: $\sqrt{x-3}-\sqrt{x-1}>0$ and  $\sqrt{4-x}-\sqrt{x}>0$. Both give you $x<2$, so the solution you obtain by squaring both members are valid only if $x<2$
When both sides are negative squaring both sides change the direction of the inequality: $-5<-1$ is true, and $25>1$, so if $x>2$ you have to change the direction of the inequality.
When the two sides have different signs you don't know what happen when you square them: $-5<1$ and $25>1$(the direction changed), but $-1<5$ and $1<25$(the direction didn't change), but in that case you already know which is bigger and you don't have to square them
A: Here a simpler way to go.
First, the domain of validity of the inequation is $[1,3]$. On this domain, rewrite it as
$$\sqrt{3-x}+\sqrt x >\sqrt{x-1}+\sqrt{4-x} .$$
Next, as both sides are nonnegative, you can square the inequation to obtain:
\begin{align}
&\phantom{\iff}\qquad 3-x+x+2\sqrt{x(3-x)} x-1+4-x+2\sqrt{(x-1)(4-x)}\\
&\iff \sqrt{x(3-x)}>\sqrt{(x-1)(4-x)} \iff x(3-x)>(x-1)(4-x) \\
&\iff  -x^2+3x >-x^2+5x-4 \iff 4>2x \quad\text{ (on the domain of validity).} 
\end{align}
