Let $X \sim \text{Bin}(n,p)$ and $t \geq 0$. I believe the following inequality holds but I don't know how to prove it:
$$ \text{Pr}[X \geq 2t] \leq \left ( \text{Pr}[X \geq t] \right )^2. $$
I have tested it numerically for many values of $n,p$ and $t$ and I found it correct. It is also "consistent" with the Chernoff Bound for the Binomial in the following sense:
Let $f(x)$ be the Chernoff upper bound for the tail $\text{Pr}[X \geq x]$, then (for appropriate $t$) it holds that $f(2t) \leq f^2(t)$.
I also believe that there should be an elegant probabilistic/combinatorial observation that gives a simple proof. Of course any help is much appreciated, I haven't succeded even in proving it via crude algebraic manipulations.
The inequality is false, at least for $t \in \mathbb{Q}^+$, take for example a Binomial($n = 10, p = 1/2$) and $t = 1/2$. I've been trying with numerical values for $t \in \mathbb{N}$, and the equality seems to hold.
Try to prove that $\text{Pr}[X < t] \leq \text{Pr}[X < 2t | X \geq t]$, and your result follows as an immediate consequence. I recommend making a sketch.