in a book published in Japan, I came across an inequality, which I am stuck at.
Let a filtered probability space $(\Omega, \mathcal{F}, \{\mathcal{F}_t\}_{t\geq 0}, P)$ given. $X=\{X_t\}_{t\geq 0}$ is a progressively measurable stochastic process with $\forall t\geq 0 ~ \mathbb{E}\left[ \int_0^t X_s^2 ds \right]<\infty$. Prove
\begin{equation} \label{eq:prob} P \left( \left| \int_0^t X(s,\omega) dB_s \right| > \alpha \right) \leq P\left( \int_0^t | X(s,\omega) |^2 ds > \beta \right) + \frac{\beta}{\alpha^2}, \end{equation}
where $\alpha, \beta, t$ are some positive constants and $\{B_t\}_{t\geq 0}$ is a standard $\mathcal{F}_t$-Brownian motion.
I somehow managed, using Chebyshev's inequality and Ito isometry, to get
\begin{align*} &P \left( \left| \int_0^T X(s,\omega) dB_s \right| > \alpha \right) = P \left( \left| \int_0^T X(s,\omega) dB_s \right|^2 > \alpha ^2 \right) \\ \leq~&\frac{1}{\alpha^2} \mathbb{E} \left[ \left( \int_0^T X(s,\omega) dB_s \right)^2 \right] = \frac{1}{\alpha^2} \mathbb{E} \left[ \int_0^T X(s)^2 ds \right] \\ =~& \frac{1}{\alpha^2} \left( \mathbb{E} \left[ \int_0^T X(s)^2 ds , \int_0^T X(s)^2 ds > \beta \right] + \mathbb{E} \left[ \int_0^T X(s)^2 ds , \int_0^T X(s)^2 ds \leq \beta \right] \right) \\ \leq~& \frac{1}{\alpha^2} \left( \beta + \mathbb{E} \left[ \int_0^T X(s)^2 ds , \int_0^T X(s)^2 ds > \beta \right] \right). \end{align*} Thus I need to show $$ \frac{1}{\alpha^2} \mathbb{E} \left[ \int_0^T X(s)^2 ds , \int_0^T X(s)^2 ds > \beta \right] \leq P\left( \int_0^T X(s)^2 ds > \beta \right). $$ This is where I am stuck.
In the first place, does \ref{eq:prob} really hold? (I searched on the web or some other textbooks and I couldn't find the above inequality.)
If so, do I need a stronger evaluation for the probability $P(|\int_0^t X ~ dB_s| > \alpha)$?
Or do I need another different approach (e.g. using sub-martingale properties of $\{ |\int_0^t X ~ dB_s| \}_t$)?