<s>[INST] Complete a proof that $F(x,y)$ is contracting. Can anyone fill in the dots in this proof?

Let $D := [0,\frac{1}{2}]^2$. Show there is exactly one $(x,y)=(x^*,y^*)\in D$ such that
  \begin{align*}
x &= \frac{x^3}{2} + y^4 + \frac{1}{4} \,, \\
y &= x^4 + \frac{y^3}{2}+\frac{1}{5} \,.
\end{align*}

Obviously I need to use Banach's fixed point theorem, so I need to show $F$ (as defined below) is contracting. 
My try (the $q \in \mathbb R$, $q<1$, isn't filled in yet of course)
Let $F(x,y) = \begin{pmatrix} \frac{x^3}{2} + y^4 + \frac{1}{4} \\ x^4 + \frac{y^3}{2}+\frac{1}{5}\end{pmatrix}$, $x=(x_1,x_2),y=(y_1,y_2)$. Then 
\begin{align*}
|F (x_1,x_2)-F(y_1,y_2)|^2 &= |F_1 (x_1,x_2)-F_1(y_1,y_2)|^2 - |F_2 (x_1,x_2)-F_2(y_1,y_2)|^2 \\
&= \left| \frac{x_1^3}{2} - \frac{y_1^3}{2} + x_2^4  - y_2^4 \right|^2 - \left|  \frac{x_2^3}{2} - \frac{y_2^3}{2} + x_1^4 - y_1^4 \right|^2 \\
&\ \ \vdots \\
&\le q^2 |x_1-y_1|^2 + q^2 |x_2 - y_2|^2 \\
&= q^2 |x-y|^2 \,.
\end{align*}
So now we have $F(x)-F(y)|\le q |x-y|$ and thus $F\colon D \to D$ contracting, $D \subset \mathbb R^2$ closed, so by Banach's fixed point theorem there exists exactly one $(x^*,y^*)\in D$ as described. $\qquad \qquad \qquad  \Box$
 [/INST] Hint: If $f$ is a continuously-differentiable real-valued function of two variables and
$$
\gamma(t) = \bigl(x_{0} + t(x - x_{0}), y_{0} + t(y - y_{0})\bigr)
$$
is the constant-speed parametrization of the segment from $(x_{0}, y_{0})$ to $(x, y)$ over $[0, 1]$, then
$$
f(x, y) - f(x_{0}, y_{0})
  = \int_{0}^{1} \nabla f\bigl(\gamma(t)\bigr) \cdot \gamma'(t)\, dt.
$$
The triangle inequality for integrals and the Cauchy-Schwarz inequality bound the absolute value in terms of the maximum of $\|\nabla f\|$ and the distance from $(x_{0}, y_{0})$ to $(x, y)$.
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