Existance and unicity of solution system I have this exercise; can anyone help me please?
$\def\arsinh{\operatorname{arsinh}}$
We have 
$$\forall x,x' \in \mathbb{R}, |\tanh (x) - \tanh (x')| \leq |x-x'|$$
and 
$$\forall x,x' \in \mathbb{R}, |\arsinh (x) - \arsinh (x')| \leq |x-x'|$$
Prove that the linear system 
$$
\begin{cases}
x - \dfrac{1}{3} \tanh (x) + \dfrac{1}{4} \arsinh (y) & = 0\\
4x - \tanh (y) + \dfrac{3}{4} \arsinh (x) & = 0
\end{cases}
$$
admits a unique solution in the space $(\mathbb{R}^2,\|\cdot\|_1)$, and determine this solution.
Note that $\arsinh$ is the “inverse hyperbolic sine”
 A: The functions $\tanh, \operatorname{arcsinh}, \operatorname{arctanh}$ are all strictly increasing, and all graphs pass through the origin.
It is trivial to verify that $(0,0)$ is a solution.
The second equation can be written as $\tanh y = 4x+\frac{3}{4} \operatorname{arcsinh} x$,
from which we get the equivalent equation $y = \operatorname{arctanh}(4x+\frac{3}{4} \operatorname{arcsinh} x )$.
Let $\phi(x) = \operatorname{arctanh}(4x+\frac{3}{4} \operatorname{arcsinh} x) $, and note that $\phi$ is also strictly increasing and $\phi(0) = 0$.
The problem now reduces to finding $x$ such that $x-\frac{1}{3} \tanh x+ \frac{1}{4} \operatorname{arcsinh} ( \phi(x) ) = 0$. It is straightforward to verify that the function $\eta(x) = x-\frac{1}{3} \tanh x+ \frac{1}{4} \operatorname{arcsinh} ( \phi(x) ) $ is strictly increasing, and $\eta(0) = 0$.
Since any $x$ that solves the original problem also satisfies $\eta(x) = 0$, we see that we must have $x=0$ at a solution. Hence $(0,0)$ is the only solution.
