Let $(Y_t)_{t \in \mathbb{N}}$ be a non-negative stochastic process and $c \in (0,1)$.
Suppose that $Y_1 = 1$ and that for all $t \in \mathbb{N}$ it holds that $$\mathbb{E}[Y_{t+1} \mid Y_1,\dots, Y_t] \le Y_t-cY_t^2.$$ Notice that this last inequality tells us that the process $(Y_t)_{t \in \mathbb{N}}$ is a (non-negative) supermartingale with the expected decreasing being at least $-c Y_t^2$.
Is it true that there exists $a,b>0$ and $T_0 \in \mathbb{N}$ such that for all $t \ge T_0$ it holds that $\mathbb{P}[Y_t > \frac{a}{t}] \le \frac{b}{\sqrt{t}}$? If it is false, what if we add the further assumptions that the process is decreasing (i.e., $Y_{t+1}\le Y_t$ for all $t \in \mathbb{N}$) and/or that $(Y_t)_{t \in \mathbb{N}}$ is a Markov chain and/or we content ourselves with the same type of bound up to polylogarithmic terms (both in the rate of shrinking and in the probability shrinking)?
What I did so far.
I realized that getting the expectation on the last inequality and using the fact that $\mathbb{E}[Y_t^2] - (\mathbb{E}[Y_t])^2 = \mathrm{Var}[Y_t] \ge 0$ we can get $\mathbb{E}[Y_{t+1}] \le \mathbb{E}[Y_{t}]-c(\mathbb{E}[Y_{t}])^2$. Defining $x_t := \mathbb{E}[Y_t]$ we get that $(x_t)_{t \in \mathbb{N}}$ is a non-negative sequence, and the recursion $x_{t+1} \le x_{t}-cx_t^2$. By induction, we can prove that $x_t \le \frac{1}{1+c(t-1)}$. In fact, this is trivial for $t=1$ since $x_1 = \mathbb{E}[Y_1] = 1$, and assuming the hypothesis holds for $t$, we have that \begin{align*} x_{t+1} &\le x_{t}-cx_t^2 = x_{t}(1-cx_t) = \frac{x_{t}(1-cx_t)(1+cx_t)}{1+cx_t} \\ &= \frac{x_{t}(1-c^2x_t^2)}{1+cx_t} \le \frac{x_{t}}{1+cx_t} \le \frac{\frac{1}{1+c(t-1)}}{1+c \frac{1}{1+c(t-1)}} \\ &= \frac{1}{1+ct} = \frac{1}{1+c\big((t+1)-1\big)} \end{align*} proving the claim by induction (we have used the fact that the function $x \mapsto \frac{x}{1+cx}$ is monotone increasing for $x \in [0,1]$).
Then, we have obtained that $\mathbb{E}[Y_t] \le \frac{1}{1+c(t-1)}$, which means that, at least in expectation, the process is going down as fast as we want.
Now, using Markov inequality we trivially get, for $\gamma > 0$ $$ \mathbb{P}\bigg[Y_t \ge \frac{1}{(1+c(t-1))^\gamma}\bigg]\le \frac{1}{(1+c(t-1))^{1-\gamma}} $$ which, for $\gamma = 1/2$, gives us a much weaker result than I hoped for in terms of the rate at which $Y_t$ shrinks to $0$ with a probability of at least $1-\frac{b}{\sqrt{t}}$.
Does anyone have any idea on how to get the improved result or maybe see a counterexample proving that the desired result is too good to be achieved?