Let $(M_n)_{n\geq0}$ be a non-negative martingale with filtration $(\mathcal{F}_n)_{n\geq0}$. Suppose $M_0=1$ and set $$T=\min\{n\geq0:M_n=0\}.$$ Also, for $R>0$, consider the stopping time $$T_R=\min\{n\geq0:M_n\geq R\}.$$ In previous parts of the question we have shown that $M_n=0$ for all $n\geq T$ almost surely, and that $\mathbb{P}(T_R<\infty)\leq 1/R$. Now suppose further that, for some constants $\sigma\in(0,\infty)$ and $\tau\in[1,\infty)$ we have $$\mathbb{E}((M_{n+1}-M_n)^2|\mathcal{F}_n)\geq\sigma^21_{\{T>n\}}\hspace{0.5cm}\text{and}\hspace{0.5cm}M_{T_R}\leq\tau R.$$ Set $Y_n=M_n^2-\sigma^2n$ and $X_n=Y_{n\wedge T\wedge T_R}$. I have shown that $X_n$ is a submartingale, and am now required to show that $$\mathbb{E}(T\wedge T_R)\leq\frac{\tau^2R}{\sigma^2}.$$
My attempt so far: For any $n\geq0$ we have $$|X_n|=|Y_{n\wedge T\wedge T_R}|=|M_{n\wedge T\wedge T_r}^2-\sigma^2(n\wedge T\wedge T_R)|\leq\tau^2R^2+\sigma^2(T\wedge T_R).$$ Thus, if I can show that $(T\wedge T_R)$ is integrable, we can use the dominated convergence theorem on $X_n$, which as $n\to\infty$ would converge almost surely to $X_{T\wedge T_R}$ (because integrability also implies that $T\wedge T_R$ is almost surely finite), and the result would follow by the fact that $\mathbb{E}(X_n)\geq\mathbb{E}(X_0)=1$ by the optional stopping theorem.
However, I can't see how to prove integrability of $(T\wedge T_R)$, so any advice and hints would be greatly appreciated!