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

You can avoid squaring by comparing $$\eqalign{ \sqrt{12}a&=\sqrt{84}+\sqrt{120}<10+11=21\ ,\cr \sqrt{12}b&=\sqrt{36}+\sqrt{228}>6+15=21\ .\cr}$$

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?

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 $a<b$