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.

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? 
