$ x^{2}- x- 4+ \frac{4\left ( x- 1 \right )^{2}}{\sqrt{3+ 2x-x^{2}}+ x+ 1}> 0 $ Solve the equation
$$x^{3}+ x+ 6= 2\left ( x+ 1 \right )\sqrt{3+ 2x-x^{2}}$$
The only answer is $x= 1$ so
I have rewritten it to
$$\left ( x- 1 \right )\left ( x^{2}- x- 4 \right )= 2\left ( x+ 1 \right ) \left (\sqrt{3+ 2x-x^{2}}- x- 1   \right )= 2\left ( x+ 1 \right )\frac{3+ 2x- x^{2}- \left ( x+ 1 \right )^{2}}{\sqrt{3+ 2x-x^{2}}+ x+ 1}= \frac{-4\left ( x+ 1 \right )^{2}\left ( x- 1 \right )}{\sqrt{3+ 2x-x^{2}}+ x+ 1}\Leftrightarrow \left ( x- 1 \right ) \left ( x^{2}- x- 4+ \frac{4\left ( x- 1 \right )^{2}}{\sqrt{3+ 2x-x^{2}}+ x+ 1} \right ) = 0$$
I need to prove $$ x^{2}- x- 4+ \frac{4\left ( x- 1 \right )^{2}}{\sqrt{3+ 2x-x^{2}}+ x+ 1}>  0 $$
Who can help me? Thanks!

Let $t=x-1$ and rewrite the equation like this
$$ t^3+3t^2+4t+8 = 2(t+2)\sqrt{4-t^2}$$
We see that $t=0$ (and so $x=1$) is a solution. Let us show it is the only one.
We also see that $t\in[-2,2]$. Now it is easy to see that $y=4t+8$ is tangent at $f(t) = t^3+3t^2+4t+8$. Let's prove that for all $t\in[-2,2]$ we have: 
$$ 2(t+2)\sqrt{4-t^2} \leq 4t+8$$
If $t=-2$ both sides are equal so let's $t>-2$. Then $t+2>0$ so we have
$$ \sqrt{4-t^2} \leq 2  $$
which is true.
Since $$f(t) > 4t+8\geq  2(t+2)\sqrt{4-t^2}$$ for each $t\in[-2,2]\setminus \{0\}$ we see that $t=0$ is the only solution.
