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: 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?
A: 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 $a<b$
A: 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.
A: You can avoid squaring by comparing
$$\eqalign{
\sqrt{12}a&=\sqrt{84}+\sqrt{120}<10+11=21\ ,\cr
\sqrt{12}b&=\sqrt{36}+\sqrt{228}>6+15=21\ .\cr}$$
A: $$a<b\iff a^2<b^2\iff 7+10+2\sqrt {70}<3+19+2\sqrt {57}$$ $$\iff 2\sqrt {70}<5+2\sqrt {57}$$ $$\iff (2\sqrt {70})^2<(5+2\sqrt {57})^2$$ $$\iff 280<25+228+20\sqrt {57}$$
$$\iff 27<20\sqrt {57}$$ 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 $2(\sqrt {70}-\sqrt {57})<2(\sqrt {81}-\sqrt {49})=4<5.$
