An application of contraction mapping theorem Suppose $\mathbb{R}$ is equipped with the standard metric and I want to show that the equation
$$
3x^3-3x^2+x-2=0
$$
has a unique root $c\geq1$. The equation is equivalent to
$$
f(x)=x,
$$
where $f(x)=1+\frac{1}{1+3x^2}$, so by showing that $f:[1,+\infty)\to \mathbb{R}$ is a contraction mapping I can prove the uniqueness of the root $c$, which will be the fixed point of $f$. Let $x,y\geq 1$, then
$$
|f(x)-f(y)|=\left|\frac{1}{1+3x^2}-\frac{1}{1+3y^2} \right|=3|x-y|\left| \frac{x+y}{(1+3x^2)(1+3y^2)}\right|
$$
So I guess I need to find a bound on $\left| \frac{x+y}{(1+3x^2)(1+3y^2)}\right|$, this is where I am stuck.

Note that $x^2-2x+1=(x-1)^2\ge0$, hence $x^2+1\ge2x$ and similarly $y^2+1\ge2y$, hence indeed (very wastefully)
$$\begin{align}(1+3x^2)(1+3y^2)&=9x^2y^2+3x^2+3y^2+1\\&\ge 9+3x^2+3y^2+1\\&=3(x^2+1)+3(y^2+1)+3\\&\ge 6x+6y+3\\&>6( x+y)  \end{align}$$
and you get a contraction factor of $\frac12$.
