$\newcommand{\diff}{\, \mathrm d} \newcommand{\and}{\quad \text{and} \quad}$Let $(\Omega, \mathcal F, \mathbb F, \mathbb P)$ be a filtered probability space where $\mathbb F = (\mathcal F_t, t \ge 0)$ is a filtration. Let $M, N$ be continuous adapted processes and $P, Q$ continuous semi-martingales. We define stochastic processes $X, Y$ by $$ X_t := \int_0^t M_s \diff P_s \and Y_t := \int_0^t N_s \diff Q_s. $$
Then $$ \langle X,Y \rangle_t = \int_0^t M_s N_s \diff \langle P, Q \rangle_s. $$
Are there some upper bounds of $|\langle X,Y \rangle_t|$ in terms of $\langle X \rangle_t$ and $\langle Y \rangle_t$?
It is mentioned at page 32 of Karatzas/Shreve's Brownian Motion and Stochastic Calculus that
5.3 Definition. For $X \in \mathscr{M}_2$, we define the quadratic variation of $X$ to be the process $\langle X\rangle_t \triangleq A_t$, where $A$ is the natural increasing process in the DoobMeyer decomposition of $X^2$. In other words, $\langle X\rangle$ is that unique (up to indistinguishability) adapted, natural increasing process, for which $\langle X\rangle_0=0$ a.s. and $X^2-\langle X\rangle$ is a martingale.
If we take two elements $X, Y$ of $\mathscr{M}_2$, then both processes $(X+Y)^2-$ $\langle X+Y\rangle$ and $(X-Y)^2-\langle X-Y\rangle$ are martingales, and therefore so is their difference $4 X Y-[\langle X+Y\rangle-\langle X-Y\rangle]$.
5.5 Definition. For any two martingales $X, Y$ in $\mathscr{M}_2$, we define their crossvariation process $\langle X, Y\rangle$ by $$ \langle X, Y\rangle_t \triangleq \frac{1}{4}\left[\langle X+Y\rangle_t-\langle X-Y\rangle_t\right] ; \quad 0 \leq t<\infty, $$ and observe that $X Y-\langle X, Y\rangle$ is a martingale. Two elements $X, Y$ of $\mathscr{M}_2$ are called orthogonal if $\langle X, Y\rangle_t=0$, a.s. $P$, holds for every $0 \leq t<\infty$.
5.7 Problem. Show that $\langle\cdot, \cdot\rangle$ is a bilinear form on $\mathscr{M}_2$, i.e., for any members $X, Y, Z$ of $\mathscr{M}_2$ and real numbers $\alpha, \beta$, we have