If you compute a continued fraction approximation $\sqrt 7 \approx \frac pq$, then for a while, you can compute $\lfloor n \sqrt 7\rfloor$ by computing $\lfloor \frac{pn}{q}\rfloor$ instead. This is very quick if we're computing the entire sequence: if we know $\hat t_n = \lfloor \frac{pn}{q}\rfloor$ and saved the value of $r_n = pn \bmod q$, then
$$
r_{n+1} = r_n + p \bmod q\qquad \text{ and } \qquad \hat t_{n+1} = t_n + \left\lfloor\frac{r_n + p}{q}\right\rfloor = \hat t_n + \frac{p + r_n - r_{n+1}}{q}.
$$
Even if you want just one term $t_n$, $\lfloor \frac{pn}{q}\rfloor$ is not too terrible to compute - it's just that computing the entire sequence $t_1, t_2, \dots, t_n$ is where the approximation really shines.
However, at some point, the approximation stops working.
In general, we can guarantee that $\lfloor \frac{pn}{q}\rfloor = \lfloor n\sqrt 7\rfloor$ for $n < q$. That's because $|\frac pq - \sqrt 7| < \frac1{q^2}$ for all continued fraction approximations, so $|\frac{pn}{q} - n\sqrt 7| < \frac{n}{q^2} < \frac1q$. As long as $\frac{pn}{q}$ is not an integer, it will be at least $\frac 1q$ away from the nearest integer, and $n\sqrt 7$ is closer than that.
So all you have to do to compute the sequence $t_n$ is to keep computing better and better continued fraction approximations to $\sqrt 7$. We have
$$
\sqrt 7 = 2 + \cfrac1{1 + \cfrac1{1 + \cfrac 1{1 + \cfrac1{4 + \dotsb}}}}
$$
and the block $1,1,1,4$ repeats forever; in continued fraction notation, $\sqrt 7 = [2; \overline{1,1,1,4}]$. There is a recursive formula for the $n^{\text{th}}$ continued fraction approximation: starting from $\frac{p_{-2}}{q_{-2}} = \frac 01$ and $\frac{p_{-1}}{q_{-1}} = \frac10$, we have $$\frac{p_n}{q_n} = \frac{a_n p_{n-1} + p_{n-2}}{a_n q_{n-1} + q_{n-2}}$$ where $a_n$ is the $n^{\text{th}}$ term of the periodic sequence $2, 1, 1, 1, 4, 1, 1, 1, 4, \dots$. To check yourself, the next few fractions you get are $$\frac 21, \frac31, \frac52, \frac83, \frac{37}{14}, \frac{45}{17}, \dots$$
Since the denominators $q_n$ grow exponentially, you will not have to do this very often if you want to compute your sequence; to compute $t_1, t_2, \dots, t_n$, you'll only need to update the continued fraction $O(\log n)$ times. (You could also begin by finding a continued fraction approximation $\sqrt 7 \approx \frac pq$ where $q$ is large enough to work for all the terms you want, then use that from the start.)
Best Answer
General algorithm:
Let us consider $d>b$. Assume $$a+\sqrt b>c+\sqrt d$$ Then $$(a-c)>(\sqrt d-\sqrt b)$$ Note that we need to proceed further, only if $a>c$, because, if $c>a$ then our assumption is certainly false. If $a>c$: $$(a-c)^2>b+d-2\sqrt {bd}$$ $$2\sqrt {bd}>b+d-(a-c)^2$$ If $(a-c)^2>b+d$, then this is certainly true, hence our assumption is true. Otherwise, we square once more, and compare. This time it's easy to compare, since all radicals are removed.
Example:
Since $27>20$, we assume:
$$4+\sqrt {20}>3+\sqrt {27}$$
We have, $1>\sqrt {27}-\sqrt {20}$
Since $4>3$, we proceed further by squaring:
$$1>47-2\sqrt {540}$$
$$2\sqrt {540}>46$$
Again, we need to proceed further, as $46>0$. So, $$\sqrt {540}>23$$ Squaring the last time, $$540>529$$ Since this is true, our original assumption was true as well.