Stability of Feedback Amplifiers
It seems that there is a logarithm sign missing. To see this, note that

$$
|A_f(j\omega)|=\left|\frac{A(j\omega)}{1+A\beta(j\omega)}\right|=\frac{|A(j\omega)|}{|1+A\beta(j\omega)|}=\frac{|A(j\omega)|}{|1+T(j\omega)|} \tag{1}\label{1}
$$
and taking the logarithm from each member of \eqref{1} you have
$$
\log |A_f(j\omega)| = \log|A(j\omega)|-\log |1+T(j\omega)|,\tag{2}\label{2}
$$
which, in my opinion is the exact form of the equation with the question mark in your notes (and is the place were the logarithm sign is missing).
Now, if 
$$
|1+T(j\omega)|&lt;1,\tag{3}\label{3}$$ 
(and not if \$|T(j\omega)|&lt;1\$, this is the place where a \$1\$ is missing) \$\log|1+T(j\omega)|&lt;0\$ and then \eqref{2} implies 

$$
\log |A_f(j\omega)| &gt; \log|A(j\omega)|\iff |A_f(j\omega)| &gt; |A(j\omega)|
$$
Now, from the point of view of the stability theory of linear systems, condition \eqref{3} means that, for a given frequency \$f\$ of the input signal \$x_i\$, the real part \$\Re[T(j\omega)]\$ of the feedback transfer function changes its sign, becoming negative. Thus, if \eqref{3} is verified, every change leading to an increase of the output signal \$x_o\$ (due not only to the increase of \$x_i\$, but also to parameter variation/drifts in \$A\$ and \$\beta\$) tends to be amplified instead of being attenuated, and this leads to the instability of the system.