For real square matrices $A, B$ of compatible dimensions, does the following inequality hold? For real square matrices $A$ and $B$ of same dimension, does this hold?
$x^{T}A^{T}B^{T}BAx\leq\|A\|^2x^{T}B^{T}Bx$ for a non-zero vector $x$? 
I know that $x^{T}A^{T}B^{T}BAx\leq\|A\|^2\|B\|^2\|x\|^2$ but I am not sure if the above inequality holds when $A$ and $B$ do not commute.
Any hints are greatly appreciated.

This is not true.
Counter Example: $$A = \begin{bmatrix}
2 & 7\\
2 & 3
\end{bmatrix}$$
$$B = \begin{bmatrix}
2 & 3\\
3 & 0
\end{bmatrix}$$
$$x = \begin{bmatrix}
2 \\
7 
\end{bmatrix}$$
