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.

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):
$3\sqrt3+\sqrt{22}<10 (9.887)$
$6\sqrt3-2\sqrt{22}>1 (1.011)$
$1+2\sqrt{22}-6\sqrt3=1-(6\sqrt3-2\sqrt{22})<0 (-0.011)$
$\sqrt11+\sqrt2<3+\sqrt3 (4.731<4.732)$
