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

• Your solution looks fine. I wouldn't waste any more time on it! Dec 10, 2019 at 18:55
• The key phrase is "without calculating the square roots" however, you are still allowed to approximate the square root which is not a clear distinction because values of irrational square roots are always approximations. There still may be clever ways to get around this. Dec 10, 2019 at 20:44
• The last inequality is $a < b$ because $a,b >0$.
– lhf
Dec 11, 2019 at 0:34
• @Ihf I can write that after, but the square root is an absolute value, so it is the same. Dec 11, 2019 at 7:58
• @Somos That phrase is wide open to interpretation, in my opinion. e.g., $\sqrt 7 \approx 2.7$, $\sqrt 10 \approx 3.2$, $\sqrt 3 \approx 1.8$, $\sqrt 19 \approx 4.4$. Although these approximations would be considered woefully imprecise for most practical purposes, I believe they would be considered a violation of the rules of the exercise. Dec 13, 2019 at 3:37

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}

• Very smart choice of $\sqrt{12}$! Dec 11, 2019 at 16:42

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 $$a; $$\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

Since $$a$$ and $$b$$ are both positive, it follows that $$a; namely $$a that is, $$a iff $$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.

$$a $$\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 $$a $$\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 $$2(\sqrt {70}-\sqrt {57})<2(\sqrt {81}-\sqrt {49})=4<5.$$