202020/202021?

This question has two parts, really. First we need to find the range of possible $a$ values; then solve $f(a)\ge0$ in this range.

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We have $b+c=7-a$

Squaring, $b^2+2bc+c^2=49-14a+a^2$

Also, $b^2+c^2=33-a^2$ ; substituting in,

$\begin{aligned} 33-a^2+2bc&=49-14a+a^2 \\ bc&=a^2-7a+8 \end{aligned}$

So $b,c$ are roots of the quadratic $x^2-(7-a)x+\left(a^2-7a+8\right)=0$

For this to have real roots, its discriminant must be non-negative; ie $(7-a)^2-4\left(a^2-7a+8\right) \ge 0$

Tidying up, this is $(a+1)(17-3a)\ge 0$

with solution $-1\le a \le\frac{17}{3}$ .

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Now we turn to solving $f(a)\ge0$ on this interval. First find the roots; we have $\sin(a-2)-\frac{202020}{202021} =0$

It's easy to check that the only roots in the interval are $a\approx3.5676$ and $a\approx 3.5739$ .

The cubes of these are, respectively, around $45.41$ and $45.65$ ; so the unique value of $\left \lfloor a^3 \right \rfloor=\boxed{45}$ .

$\begin{cases} a+b+c=7 && (1) \\ \\ a^2+b^2+c^2=33 &&(2) \\ \\ \sin(a-2)\ge \dfrac{202020}{202021 } && (3) \end{cases}$ Using equation $(1)$ and $(2)$ , $\begin{aligned} (a+b+c)^2&=7^2 \\ a^2+b^2+c^2+2a(b+c)+2bc&=49 \\ 33+2a(7-a)+2bc&=49 \\ 14a-2a^2+2bc&=16 \end{aligned}$ In equation $(2)$ , applying the AM-GM inequality, $\begin{aligned} \frac{b^2+c^2}{2}\ge bc&\implies 33-a^2\ge 2bc \\ &\implies 33-a^2\ge 16-(14a-2a^2) \\ &\implies -3a^2+14a+17\ge 0 \\ &\therefore \; -1\le a\le \frac{17}{3} \end{aligned}$ Combining this with equation $(3)$ , $\begin{aligned} \sin^{-1} \left(\frac{202020}{202021 }\right)+2&\le a\le 2+\pi-\sin^{-1} \left(\frac{202020}{202021 }\right) \\ \implies 3.5676...&\le a\le 3.5739... \\ \implies 45.407...&\le a \le 45.648... \end{aligned}$ Therefore, the only possible value of $\lfloor a^3\rfloor$ is $\boxed{45}$