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

Question

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):

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

Answer 1

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?

Answer 2

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.

Answer 3

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

Answer 4

$$a<b\iff a^2<b^2\iff 7+10+2\sqrt {70}<3+19+2\sqrt {57}$$ $$\iff 2\sqrt {70}<5+2\sqrt {57}$$ $$\iff (2\sqrt {70})^2<(5+2\sqrt {57})^2$$ $$\iff 280<25+228+20\sqrt {57}$$ $$\iff 27<20\sqrt {57}$$ 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 $2(\sqrt {70}-\sqrt {57})<2(\sqrt {81}-\sqrt {49})=4<5.$

Answer 5

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 $a<b$