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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.

Blue
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Pradeep Suny
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  • Assume $y\leq x$. Then $y^2\leq x^2$. This new inequality only contains the roots $\sqrt 22$ and $\sqrt 3$. Isolate one root, square the expression again. Now isolate the remaining root and obtain a contradiction. – Zuy Jan 11 '22 at 14:30

2 Answers2

9

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. $$ \begin{align}\sqrt{11}+\sqrt{2} \qquad &?\qquad 3+\sqrt{3},\\ 11 + 2\sqrt{22} + 2 \qquad &?\qquad 9+6\sqrt{3}+3,\\ 1 + 2\sqrt{22} \qquad &?\qquad 6\sqrt{3},\\ 1 + 4\sqrt{22} + 88 \qquad &?\qquad 108,\\ 4\sqrt{22} \qquad &?\qquad 19,\\ 352 \qquad &?\qquad 361. \end{align} $$

ACB
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Vasily Mitch
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0

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):

$3\sqrt3+\sqrt{22}<10 (9.887)$

$6\sqrt3-2\sqrt{22}>1 (1.011)$

$1+2\sqrt{22}-6\sqrt3=1-(6\sqrt3-2\sqrt{22})<0 (-0.011)$

$\sqrt11+\sqrt2<3+\sqrt3 (4.731<4.732)$

Oscar Lanzi
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