So let me begin by stating the matrix deviation inequality (Theorem 9.1.1) in Vershynin's High Dimensional Probability book.
Theorem 9.1.1: We have that if $A$ is an $m\times n$ matrix whose rows $A_i$ are independent, isotropic, and sub-gaussian random vectors in $\mathbb{R}^n$. Then for any subset $T\subset \mathbb{R}^n$, we have
$$E\sup_{x\in T} \left| |||Ax||_2 - \sqrt{m} ||x||_2 \right | \leq CK^2 \gamma(T)$$
where $\gamma(T)=E\sup_{t\in T}|\langle x,g\rangle|$ where $g \sim \mathcal{N}(0, I_n)$ is the Gaussian complexity and $K=max_i ||A_i||_{\psi_2}$.
Now under the same assumptions as Theorem 9.1.1 exercise 9.1.10 asks us to extend the result to squares. Namely, show that
$$E\sup_{x\in T} \left| |||Ax||_2^2 - m ||x||_2^2 \right | \leq CK^4 \gamma(T)^2+CK^2 \sqrt{m} \text{rad}(T)\gamma(T)$$
where $\text{rad}(T)=\sup_{t\in T} ||t||_2$
The idea of proof I had in mind was to rewrite the expectation as
$$ \begin{align*} E\sup_{x\in T} \left| |||Ax||_2^2 - m ||x||_2^2 \right| &= E\sup_{x\in T} \left(\left| |||Ax||_2 - \sqrt{m} ||x||_2 \right| \left| |||Ax||_2 + \sqrt{m} ||x||_2 \right|\right) \\ & \leq E\sup_{x\in T} \left(\left| |||Ax||_2 - \sqrt{m} ||x||_2 \right| \left| |||Ax||_2 + \sqrt{m} ||x||_2 \right|\right) \\ & \leq E\sup_{x\in T} \left(\left| |||Ax||_2 - \sqrt{m} ||x||_2 \right| \left| |||Ax||_2 \right|\right) + E\sup_{x\in T} \left(\left| |||Ax||_2 - \sqrt{m} ||x||_2 \right| \left|\right|\right) \sqrt{m} ||x||_2 \\ & \leq E\sup_{x\in T} \left(\left| |||Ax||_2 - \sqrt{m} ||x||_2 \right| \left| |||Ax||_2 \right|\right) + CK^2\sqrt{m}\text{rad}(T)\gamma(T) \end{align*} $$
My issue is how do I show $E\sup_{x\in T} \left(\left| |||Ax||_2 - \sqrt{m} ||x||_2 \right| \left| |||Ax||_2 \right|\right) \leq CK^4 \gamma(T)^2$