Without calculating the square roots, determine which of the numbers:$a=\sqrt{7}+\sqrt{10}\;\;,\;\; b=\sqrt{3}+\sqrt{19}$ is greater. 
Without calculating the square roots, determine which of the numbers:
$$a=\sqrt{7}+\sqrt{10}\;\;,\;\; b=\sqrt{3}+\sqrt{19}$$
is greater.

My work (I was wondering if there are other ways to prove this):
$$a^2=17+2\sqrt{70}, \;\;b^2=22+2\sqrt{57}$$
$$\sqrt{64}<\sqrt{70}<\sqrt{81}\implies 8<\sqrt{70}<9\implies a^2<35$$
$$\sqrt{49}<\sqrt{57}<\sqrt{64}\implies 7<\sqrt{57}<8\implies b^2>36$$
$$a^2<35<36<b^2\implies a^2<b^2\implies |a|<|b|$$

There are indeed other ways to do this. Your solution is great, but if you were just curious about another method, here is one:
$$
\begin{align}
\sqrt{7} + \sqrt{10} \quad &? \quad \sqrt{3} + \sqrt{19} \\
\sqrt{10} - \sqrt{3} \quad &? \quad \sqrt{19} - \sqrt{7}
\end{align}
$$
Note that instead of comparing $a$ and $b$ directly, we can just compare these values.
Define the function
$$
f(x) = \sqrt{9x+10} - \sqrt{4x+3}
$$
We do this because $f(0) = \sqrt{10} - \sqrt{3}$ and $f(1) = \sqrt{19} - \sqrt{7}$.
The magic step is now figuring out that for all positive $x$, this function is increasing, which tells us that $f(1) > f(0)$.
Of course, seeing that this function is increasing is not exactly obvious, but it is not a difficult task if you have a calculus background.
Perhaps there is another step we can take or a different function we can use that would make the fact it is increasing more obvious?
