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):
$$a^2=17+2\sqrt{70}, \;\;b^2=22+2\sqrt{57}$$
$$\sqrt{64}<\sqrt{70}<\sqrt{81}\implies 8<\sqrt{70}<9\implies a^2<35$$
$$\sqrt{49}<\sqrt{57}<\sqrt{64}\implies 7<\sqrt{57}<8\implies b^2>36$$
$$a^2<35<36<b^2\implies a^2<b^2\implies |a|<|b|$$

Another way of doing it. Using the result that,
$$\sqrt{1+\frac{n}{m}}-\sqrt{1-\frac{n}{m}}  ≥ \frac{n}{m}$$
$$\sqrt{1+\frac{3}{32}}-\sqrt{1-\frac{3}{32}}  >\frac{3}{32}$$
$$\sqrt{1+\frac{3}{32}} > \frac{3}{32}+\sqrt{1-\frac{3}{32}}$$
$$8\sqrt{1+\frac{3}{32}} > \frac{3}{4}+8\sqrt{1-\frac{3}{32}}$$
$$\sqrt{70} > \frac{3}{4}+\sqrt{57}\tag{1}$$
Assuming $a<b$;
$$\sqrt{7}+\sqrt{10}<\sqrt{3}+\sqrt{19}$$
$$17+2\sqrt{70}<22+2\sqrt{57}$$
$$\sqrt{70}<\frac{5}{2}+\sqrt{57}$$
By Using (1);
$$\sqrt{70}<\frac{5}{2}+\sqrt{70}-\frac{3}{4}$$
$$0<1+\frac{3}{4}$$
Hence true, so $a<b$
