<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] In more detail they'll you'll probably need.
For $\sqrt{3 - x}$ to exist at all in the reals it must be $3 - x \ge 0$ so $x \le 3$.  For $\sqrt{x+1}$ to exist in the reals it must be $x + 1 \ge 0$ so $x \ge -1$.  So $-1 \le x \le 3$.  Those are the only values where the term $\sqrt{3 - x} - \sqrt{x+1}$ can make any sense.  
$\sqrt{3-x} - \sqrt{x+1} > 1/2$
We want to remove the square roots.  To do that we will have to square them.  
There are three things we have to think about when we square sides. 
i) $(a + b)^2 = a^2 + 2ab + b^2$ has three terms and the middle term still involves $a$ and $b$, so $(\sqrt a + \sqrt b)^2 = a + 2\sqrt a \sqrt b + b$ will still have the roots combined.  So we probably want to isolate the roots to each side of the inequality.  If $\sqrt a + \sqrt b = c$ then $\sqrt a = c - \sqrt b$ so $\sqrt a ^2 = (c - \sqrt b)^2$ so $a = c^2 - 2c\sqrt b + b$.  We still have a square root but only one.  So we have made progress.
ii) We need to be concerned about signs.  If $a > b >0$ than $a^2 > b^2$.  But if $ 0 > a > b$ then the opposite is true.  We must "flip" the inequality sign.  If $0 > a > b$ then $a^2 < b^2$.  If $a > 0 > b$ then we don't know if $a^2 > b^2$ or $a^2 = b^2$ or $a^2 < b^2$.  Any one of those may be true.
iii) We need to be aware of "extraneous solutions".  If we have $x > \sqrt{ x - 5}$ we know $x$ is positive. If  we square to get $x^2 > x - 5$ we have now added the extraneous possibility that $x$ might be negative.  But we know it isn't.  We must take that into account.
So let's do this:
$\sqrt{3 -x} - \sqrt{x+1} > 1/2$
Let's move one of the radicals to the other side:
$\sqrt{3 - x} > 1/2 + \sqrt{x+1}$
We know $1 > 0$ and $\sqrt{x+1} \ge0$ so if we square both sides they will be positive and the inequality signs will remain as they are.
$\sqrt{x- 3}^2 > (1/2 + \sqrt{x + 1})^2$
$3 - x > 1/4 + \sqrt{x + 1} + x + 1 = x + 1 1/4 + \sqrt{x+1}$.
Note: we have just added the extraneous possibility that $x > 3$ or $x < -1$.  But we already made a note that those will be impossible.
We still have a square root to isolate:
$3 - x > x + 1 1/4 + \sqrt{x+1}$ so
$7/4 - 2x > \sqrt{x+1}$; $ \sqrt{x+1} \ge 0$ so we can square and keep the "$>$" unflipped
$(7/4 - 2x)^2 > \sqrt{x + 1}^2$
$49/16 - 7x + 4x^2 >(x + 1)$
Note: By squaring we have added the extraneous possibility that $7/4 - 2x$ might be negative.  We know that isn't true.  So we must note: $7/4 - 2x \ge 0$ so $7/8 \ge x$ . We also know $-1 \le x \le 3$ so combining those we know $-1 \le x \le 7/8$.
Let's go on:
$49/16 - 7x + 4x^2 >(x + 1)$
$33/16 - 8x + 4x^2 >0$
$33/64 - 2x + x^2 >0$
Use the quadratic equation:
$(x - \frac{2 + \sqrt{4 +33/16}}{2})(x - \frac{2 + \sqrt{4 +33/16}}{2}) > 0$
$(x - 1-\frac{\sqrt{33}}{8})(x - 1 + \frac{\sqrt{31}}{8}) > 0$
We have two terms multiplying to  a positive result.  So EITHER they are both positive or the are both negative.
If they are both positive we have:
$x > 1 + \frac{\sqrt{31}}{8}$
But we know $-1 \le x \le 7/8$ so this is impossible.
So they are both negative.  So we know:
$x < 1 - \frac{\sqrt{31}}{8} < 1 + \frac{\sqrt{31}}{8}$
$1 - \frac{\sqrt{31}}{8} \approx .304$ so $-1 < 1 - \frac{\sqrt{7}}{2} < 7/8$
We know $-1 \le x \le 7/8$ and $x < 1- \frac{\sqrt{31}}{8}$. 
So combining we know:
$-1 \le x < 1 - \frac{\sqrt{31}}{8}$.
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