<s>[INST] 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):
$$a^2=17+2\sqrt{70}, \;\;b^2=22+2\sqrt{57}$$
$$\sqrt{64}<\sqrt{70}<\sqrt{81}\implies 8<\sqrt{70}<9\implies a^2<35$$
$$\sqrt{49}<\sqrt{57}<\sqrt{64}\implies 7<\sqrt{57}<8\implies b^2>36$$
$$a^2<35<36<b^2\implies a^2<b^2\implies |a|<|b|$$
 [/INST] Since $a$ and $b$ are both positive, it follows that $a<b \iff a^2<b^2$; namely $$a<b\iff17+2\sqrt{70}<22+2\sqrt{57};$$ that is, $a<b$ iff $2\left(\sqrt{70}-\sqrt{57}\right)<5.$ Continuing equivalent statements in this way, we get
$$
a < b
 \iff 508-8\sqrt{3990}<25
 \iff 60 \tfrac38 < \sqrt{3990}
 \iff 3600+45+\tfrac9{64} < 3990,
$$
the latter clearly being the case.
</s>