If $\small {x+\sqrt { (x+1)(x+2) } +\sqrt { (x+2)(x+3) } +\sqrt { (x+3)(x+1) } = 4}$, solve for $x$. I came across this olympiad algebra problem, asking to solve for $x$:
$x\ +\ \sqrt { (x+1)(x+2) } \ +\ \sqrt { (x+2)(x+3) } +\ \sqrt { (x+3)(x+1) } =\ 4$
Here was my try:
If $$x\ +\ \sqrt { (x+1)(x+2) } \ +\ \sqrt { (x+2)(x+3) } +\ \sqrt { (x+3)(x+1) } =\ 4$$
Then $\quad \sqrt { (x+1)(x+2) } +\sqrt { (x+2)(x+3) } +\sqrt { (x+3)(x+1) } =4-x$.
Further, I tried squaring the equation on both sides, but that doesn't seem to solve my problem. Please help.
Thank you.
 A: Hint: Observe that $:x = (x+1) + (x+2) - (x+3)$. This leads us to letting $a = \sqrt{x+1}, b = \sqrt{x+2}, c = \sqrt{x+3} \Rightarrow a^2+b^2-c^2 + ab+bc+ca = 4, b^2-a^2 = 1 = c^2-b^2 \Rightarrow a^2-1 +ab+bc+ca= 4 \Rightarrow (a+b)(a+c) = 5 \Rightarrow \dfrac{a+c}{b-a} = 5 \Rightarrow a+c = 5b-5a\Rightarrow 6a = 5b-c\Rightarrow 6\sqrt{x+1} = 5\sqrt{x+2} - \sqrt{x+3}$. Can you continue? 
A: While Kf-Sansoo has given an elegant answer, if the problem asks for any real (and not just rational) solution, then it misses a second one which is a root of a quartic equation, hence normally is not easy to do by hand.
In general, the two solutions to,
$$x+\sqrt{(x+1)(x+2)}+\sqrt{(x+2)(x+3)}+\sqrt{(x+3)(x+1)} = n\tag1$$
for real $n>0$ are,
$$x = \frac{(n^2+4n+5)^2}{4(n+1)(n+2)(n+3)}-2\tag{2a}$$ 
and the appropriate root of,
$$-23 + 48 n - 22 n^2 + n^4 - 4 (30 - 33 n + 6 n^2 + n^3) x \\+ 
  16 (-11 + 6 n) x^2 + 16 (-6 + n) x^3 - 16 x^4=0\tag{2b}$$
For $n = 4$, we have $x_1 = -311/840 \approx -0.37$. Then $x_2 \approx -5.12357$ as a root of,
$$73 - 232 x + 208 x^2 - 32 x^3 - 16 x^4 = 0$$
with both valid for the positive case of $\sqrt{z}$ as the graph from Walpha below shows,


$\color{green}{Edit:}$ When using Kf-Sansoo's method, we end up with an expression of form,
$$\prod^4 (c_1\sqrt{x+1}\pm c_2\sqrt{x+2}\pm c_3\sqrt{x+3}) = 0$$
Let $n=4$ and we get $x = -311/840$. Simpler, but the price to pay is we lose a second solution. Another method is to form an octic,
$$\prod^8 \big(y-(\pm\sqrt{z_1}\pm \sqrt{z_2}\pm \sqrt{z_3})\big)=0 \tag3$$
After it is formed, substitute into $(3)$ the ff,
$$y = n-x\\z_1=(x+1)(x+2)\\z_2=(x+2)(x+3)\\z_3=(x+3)(x+1)$$
and we get linear/quartic factors given by $(2a), (2b)$. Less simpler, but it yields a second valid solution.
