Show $1+a+b+c\leq 2\lfloor\sqrt{at^2+bt+c}\rfloor$ for all $a,b,c,t\in\Bbb N\cup\{0\}$ with $a\neq 0$, $t\ge 2$ and $a,b,c\le t-1$. Related to a project I'm writing, I came across the problem of showing that 
$$1+a+b+c \leq 2\lfloor \sqrt{ at^2+bt+c}\rfloor$$
holds for all nonnegative integers $a,b,c,t$ satisfying $a \neq 0$, $t \geq 2$ and $a,b,c \in \{0,1,\ldots,t-1\}$. 
I have checked numerically that it holds true for all valid choices of $a,b,c$ for $t\leq 300$, but I'm not able to prove that the statement is always true. I'm suspecting that there is some simple argument to why, but unfortunately I don't see it. Can anyone help ? :) 
 A: You're asking to confirm
$$1+a+b+c \leq 2\lfloor \sqrt{ at^2+bt+c}\rfloor \tag{1}\label{eq1A}$$
Since $a,b,c \in \{0,1,\ldots,t-1\}$, each value is at most $t - 1$, so you get
$$1 + a + b + c \le 1 + (t - 1) + (t - 1) + (t - 1) = 3t - 2 \tag{2}\label{eq2A}$$
With $a \ge 4$, you have
$$2\lfloor \sqrt{ at^2+bt+c}\rfloor \ge 2\lfloor \sqrt{ at^2}\rfloor \ge 2(2t) = 4t \gt 3t - 2 \tag{3}\label{eq3A}$$
This shows \eqref{eq1A} always holds for $a \ge 4$, so you only need to check $a = 1$, $2$ and $3$. Consider first $a = 1$. Then
$$1 + a + b + c \le 1 + 1 + (t - 1) + (t - 1) = 2t \tag{4}\label{eq4A}$$
Since for $a = 1$, you have $t^2 + bt + c \ge t^2$, you thus have $2\lfloor \sqrt{ at^2+bt+c}\rfloor \ge 2(t)$, so \eqref{eq1A} holds.
For $a = 2$, you have $1 + a + b + c \le 2t + 1$. However, if $b \lt t - 1$ or $c \lt t - 1$, then $1 + a + b + c \le 2t$, so the condition will hold since $2t^2 \gt t^2$ which we've shown above works. Thus, only need to check $2t^2 + (t - 1)t + (t - 1) = 3t^2 - 1 \ge (t + 1)^2 = t^2 + 2t + 1$. This is true since with $t \ge 2$, then $t^2 \ge 2t$, so $2t^2 \ge 4t$. Thus, $3t^2 - 1 = t^2 + 2t^2 - 1 \ge t^2 + 4t - 1 \gt t^2 + 2t + 1$.
I'll leave it to you to confirm \eqref{eq1A} also holds for $a = 3$ (hint: $\sqrt{3}t \gt t + 1$ for $t \ge 2$).
Update: The $a = 2$ and $a = 3$ cases can be combined by using that $t \ge a + 1$. With $a \le 3$, you have $1 + a + b + c \le 1 + (t - 1) + (t - 1) + 3 = 2t + 2 = 2(t + 1)$. Also, $(\sqrt{a})t \gt t + 1$, so you can use an argument similar to the one in this answer.
A: For the case $a=3$ we have: 
$$a+b+c+1=4+b+c \leq 4+2(t-1)=2t+2$$
We also have that:
$$ 2 \lfloor  \sqrt{at^2+bt+c} \rfloor = 2 \lfloor \sqrt{3t^2+bt+c} \rfloor  \geq 2 \lfloor \sqrt{3t^2} \rfloor \geq 2 \lfloor t+1\rfloor=2t+2$$
Where we used that $\sqrt{3t^2} > (t+1)$ for all integers $t \geq 2$.
Therefore, it holds also for $a=3$
