Which is bigger, $1+3\sqrt{2}$ or $3\sqrt{3}$? Which is bigger, $a = 1+3\sqrt2$ or $b = 3\sqrt3$?
To find out result, I am doing: $(3\sqrt3)^2-(1+3\sqrt2)^2=8-2\sqrt{18}$.
But how can I find if the value of $8-2\sqrt{18}$ is positive or negative?
Thank you.
 A: Hint:
$$\color{blue}8-\color{red}{2\sqrt{18}}=\color{blue}{\sqrt{64}}-\color{red}{\sqrt{4}\sqrt{18}}=\color{blue}{\sqrt{64}}-\color{red}{\sqrt{72}}$$
A: HINT:
Note that $(3\sqrt3)^2-(1+3\sqrt2)^2=27-1-6\sqrt2-18=8-6\sqrt2$. Next note that $\sqrt2>\frac86$.
A: Hint
$$8^2=64 <72=(2\sqrt{18})^2$$
A: You can use an “unknown relation”: denote by ? either $<$ or $>$; so long as you do operations that don't change the direction of inequalities, the symbol will always mean the same; so, when you're done you'll know which one it is.
\begin{align}
1+3\sqrt{2}&\mathrel{?}3\sqrt{3}\\
(1+3\sqrt{2})^2&\mathrel{?}(3\sqrt{3})^2\\
1+6\sqrt{2}+18&\mathrel{?}27\\
6\sqrt{2}&\mathrel{?}27-1-18\\
6\sqrt{2}&\mathrel{?}8\\
3\sqrt{2}&\mathrel{?}4\\
(3\sqrt{2})^2&\mathrel{?}16\\
18&\mathrel{?}16
\end{align}
So ? was $>$.
A: Alternatively $\rm\ \dfrac{2}{a-5} = \color{#c00}4+\color{#0a0}{3\sqrt2}\, <\, \color{#c00}5+\color{#0a0}{3\sqrt3} = \dfrac{2}{b-5}\ $ since $\rm\, \color{#c00}{4<5},\,\ \color{#0a0}{3\sqrt2 < 3\sqrt 3} $ 
Flipping gives $\rm\ a-5\, >\, b-5 \ \Rightarrow\ a > b\ \ $ QED
Remark $\ $ This is not pulled out of a hat. Rather, it arises from the continued fraction algorithm for comparing real numbers. Namely, first we compare the integer parts of $\rm\,a,b.\,$ Both are easily seen to be $\,5,\,$ so we subtract $5$ from both to get their fractional parts $\rm\,a-5,\ b-5,\,$ then recurse, comparing their reciprocals $\rm\,1/(a-5),\ 1/(b-5),\,$ etc, etc.
