Comparing $\frac {9}{\sqrt{11} - \sqrt{2}}$ and $\frac {6}{3 - \sqrt{3}}$ (without calculator) We want to compare the following two numbers:
$$x = \frac {9}{\sqrt{11} - \sqrt{2}} \quad\text{and}\quad y = \frac {6}{3 - \sqrt{3}}$$
My attempts so far:
I multiply both numerator and denominator of $x$ by $\sqrt{11} + \sqrt{2}$ so I get:
$$x = \frac {9(\sqrt{11} + \sqrt{2})}{(\sqrt{11} - \sqrt{2})\cdot(\sqrt{11} + \sqrt{2})}$$ so
$$x = \frac {9(\sqrt{11} + \sqrt{2})}{(11 - 2)} = \sqrt{11} + \sqrt{2}$$
Similarly,  $y = 3 + \sqrt{3}$.
But how do I take it from this point forward?
Of course $y>x$ but I must prove it.
I also tried to compare $\sqrt 11$ with $\sqrt 12$ which equals $2 \sqrt3$ but again I am not getting anywhere.
Thank you.
 A: Not only you can multiply number by the same positive value, you can also add or subtract the same values. The general idea how to solve similar questions is to remove roots one by one (making sure that values to be squared are positive):
$$
a+\sqrt{p}+\sqrt{q}+\ldots+\sqrt{r} \qquad ?\qquad  b+\sqrt{u}+\sqrt{v}+\ldots+\sqrt{w},\\
a+\sqrt{p}+\sqrt{q}+\ldots+\sqrt{r}-b-\sqrt{v}-\ldots-\sqrt{w} \qquad ?\qquad  \sqrt{u},\\
(a+\sqrt{p}+\sqrt{q}+\ldots+\sqrt{r}-b-\sqrt{v}-\ldots-\sqrt{w})^2 \qquad ?\qquad  u,
$$
rinse and repeat.
However, in your case, since the left side has only 2 roots, we can square first, to diminish the number of roots straight away.
$$
\begin{align}\sqrt{11}+\sqrt{2}  \qquad &?\qquad  3+\sqrt{3},\\
11 + 2\sqrt{22} + 2 \qquad &?\qquad  9+6\sqrt{3}+3,\\
1 + 2\sqrt{22} \qquad &?\qquad  6\sqrt{3},\\
1 + 4\sqrt{22} + 88 \qquad &?\qquad  108,\\
4\sqrt{22} \qquad &?\qquad  19,\\
352 \qquad &?\qquad  361.
\end{align}
$$
A: Render
$[(\sqrt{11}+\sqrt2)-(3+\sqrt3)][(\sqrt{11}+\sqrt2)-(3+\sqrt3)]=(\sqrt{11}+\sqrt2)^2-(3+\sqrt3)^3=(13+2\sqrt{22})-(12+6\sqrt3)=1+2\sqrt{22}-6\sqrt3$
Then
$6\sqrt3-2\sqrt{22}=2(3\sqrt3-\sqrt{22})=\dfrac{2(27-22)}{3\sqrt3+\sqrt{22}}=\dfrac{10}{3\sqrt3+\sqrt{22}}$
And finally
$(3\sqrt3+\sqrt{22})^2=49+6\sqrt{66}<49+6\sqrt{8×9}\overset{AM-GM}{<}49+(6×8.5)=100$
So (parentheses indicate calculator results):
$3\sqrt3+\sqrt{22}<10 (9.887)$
$6\sqrt3-2\sqrt{22}>1 (1.011)$
$1+2\sqrt{22}-6\sqrt3=1-(6\sqrt3-2\sqrt{22})<0 (-0.011)$
$\sqrt11+\sqrt2<3+\sqrt3 (4.731<4.732)$
