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

$$a<b\iff a^2<b^2\iff 7+10+2\sqrt {70}<3+19+2\sqrt {57}$$ $$\iff 2\sqrt {70}<5+2\sqrt {57}$$ $$\iff (2\sqrt {70})^2<(5+2\sqrt {57})^2$$ $$\iff 280<25+228+20\sqrt {57}$$
$$\iff 27<20\sqrt {57}$$ and we have $20\sqrt {57}>20\sqrt 4=40>27.$ 
Another way, from one of the intermediate steps above, is $$a<b\iff 2\sqrt {70}<5+2\sqrt {57}$$ $$\iff 2(\sqrt {70}-\sqrt {57})<5$$ $$\iff 2(\sqrt {70}-\sqrt {57})(\sqrt {70}+\sqrt {57})<5(\sqrt {70}+\sqrt {57})$$ $$\iff 2(70-57)<5(\sqrt {70}+\sqrt {57})$$ and we have $5(\sqrt {70}+\sqrt {57})>5\sqrt {70}>5\sqrt {64}=40>26=2(70-57).$
Or we might notice that $2(\sqrt {70}-\sqrt {57})<2(\sqrt {81}-\sqrt {49})=4<5.$
