Let $A,B\in\mathcal{M}_{n\times n}(\mathbb{R})$ be matrices such that $\|A\|_{F}\cdot\|B\|_{F}<\frac{9}{16}$, where $\|\cdot\|_F$ denotes the Frobenius norm. Show that in $B[0_{n\times n},1/2]$ (closed ball) the equation:
$$AX^{2}B-\frac{1}{3\sqrt{n}}I=X$$
has a solution using the Fixed Point Theorem.
I defined the aplication $X\mapsto\varphi(X)=AX^{2}B-\frac{1}{3\sqrt{n}}I$ to use the fixed point theorem. The ball is closed and I already prove that $B[0_{n\times n},1/2]\subseteq\varphi(B[0_{n\times n},1/2])$, then I just need to show that it is a contracting function:
Let $X,Y\in B[0_{n\times n},1/2]$, then:
$\|\varphi(X)-\varphi(Y)\|_F=\|AX^{2}B-AY^{2}B\|_F\leq\|A\|_{F}\|X^2-Y^2\|_F\|B\|_F<\frac{9}{16}\|X^2-Y^2\|_F$
I don't know how to make appear $\|X-Y\|_F$. Any hint?
$$ \begin{aligned} &\|AX^2B-AY^2B\|_F\\ &=\|A\,\left[\,X(X-Y)+(X-Y)Y\,\right]\,B\|_F\\ &\le\|A\|_F\|B\|_F\,\left(\|X\|_F+\|Y\|_F\right)\,\|X-Y\|_F\\ &\le\left(\frac{9}{16}\right)^2\left(\frac{1}{2}+\frac{1}{2}\right)\|X-Y\|_F. \end{aligned} $$