Comparing the magnitudes of expressions of surds

Question

I recently tackled some questions on maths-challenge / maths-aptitude papers where the task was to order various expressions made up of surds (without a calculator, obviously).

I found myself wondering whether I was relying too much on knowing the numerical value of some common surds, when a more robust method was available (and would work in more difficult cases).

For example, one question asked which is the largest of:

(a) $\sqrt{10}$
(b) $\sqrt2+\sqrt3$
(c) $5-\sqrt3$

In this case, I relied on my knowledge that $\sqrt{10} \approx 3.16$ and $\sqrt2\approx 1.41$ and $\sqrt3 \approx 1.73$ to find (a) $\approx 3.16$, (b) $\approx ~3.14$ and (c) $\approx ~3.27$ so that the required answer is (c).

But this seemed inelegant: I felt there might be some way to manipulate the surd expressions to make the ordering more explicit. I can't see what that might be, however (squaring all the expressions didn't really help).

I'd appreciate some views: am I missing a trick, or was this particular question simply testing knowledge of some common values?

EDIT: after the very helpful answers, which certainly showed that there was a much satisfying and general way of approaching the original question, can I also ask about another version of the question which included (d) $\sqrt[4]{101}$.

When approaching the question by approximation, I simply observed that $\sqrt[4]{101}$ is only a tiny bit greater than $\sqrt{10}$, and hence it still was clear to choose (c) as the answer. Is there any elegant way to extend the more robust methods to handle this case?

Answer 1

You can use:

(1) the fact that $f(x)=x^2$ is a monotonically increasing function when $x\geq0$ and

(2) the arithmetic-geometric mean inequality $\sqrt{ab}\leq\frac{a+b}{2}$, when $a, b\geq0$. Hence, $$ (\sqrt{2}+\sqrt{3})^2=5+2\sqrt{2\cdot3}\leq5+2\frac{2+3}{2}=5+5=10=(\sqrt{10})^2 $$ Therefore, using (1), we obtain $\sqrt{2}+\sqrt{3}\leq 10$. I forgot about this: $$ 5-\sqrt{3}=3+2-\sqrt{3}=3+\frac{1}{2+\sqrt{3}}\geq3+\frac{1}{2+2}=3.25 $$ One can easily verify that $(3+1/4)^2>10.5>10$. One also finds that $10.5^2>110>101$.

Then, performing argument (1) twice, one finds that $5-\sqrt{3}>(101)^{1/4}$.

Consequently, $5-\sqrt{3}$ is the bigger number.

Answer 2

Comparing $\sqrt{10}$ and $\sqrt2+\sqrt3$ is the same as comparing $10$ and $(\sqrt2+\sqrt3)^2=5+2\sqrt6$. That's the same as comparing $5$ and $2\sqrt6$. Which of these is bigger?

Likewise comparing $\sqrt{10}$ and $5-\sqrt3$ is the same as comparing $10$ and $(5-\sqrt3)^2=28-10\sqrt3$. That's the same as comparing $10\sqrt3$ and $18$.

Which of these is bigger?