Difference of Square root and order relation

Question

If $a=\sqrt{15} -\sqrt{11}$, $b=\sqrt{27} - \sqrt{23}$, $c=\sqrt{6} - \sqrt{5}$ then which of the following is correct?

A. a>b>c
B. a>c>b
C. c>b>a
D. c>a>b

I am new to number system and don't want to calculate the square root to find out the answer. Is there any easy way to solve this?

Please help.

Answer 1

Hint: Use the fact that $\sqrt{x}-\sqrt{x-d}=\frac{d}{\sqrt{x}+\sqrt{x-d}}$.

So we have $$a=\frac{4}{\sqrt{15}+\sqrt{11}}, b=\frac{4}{\sqrt{27}+\sqrt{23}}$$ Clearly $a>b$. As for $c$, although we have $d=1$, we can multiply both top and bottom by $4$ (for a like for like comparison), resuting in $$c=\frac{1}{\sqrt{6}+\sqrt{5}}=\frac{4}{\sqrt{96}+\sqrt{80}}$$ which (given the sum of its denominator) is by far the smallest.

Answer 2

The mean value theorem gives us that $\sqrt{a}-\sqrt{b}=\frac{a-b}{2\sqrt{c}}$ for some $c$ between $a,b$. This lets you compare the first two, easily, since $a-b$ is the same for both cases.