Why rationalize the denominator?
This was very important before computers in problems where you had to do something else after computing an answer.
One simple example is the following: When you calculate the angle between two vectors, often you get a fraction containing roots. In order to recognize the angle, whenever when possible, it is good to have a standard form for these fractions [side note, I saw often students not being able to find the angle $\theta$ so that $\cos(\theta)=\frac{1}{\sqrt{2}}$]. The simplest way to define a standard form is by making the denominator or numerator integer.
If you wonder why the denominator is the choice, it is the natural choice: As I said often you need to make computations with fractions. What is easier to add: $$\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{6}+\sqrt{3}} \, \mbox{ or }\, \frac{\sqrt{3}}{3}+\frac{\sqrt{6}-\sqrt{3}}{3} \,?$$
Note that bringing fractions to the same denominator is usually easier if the denominator is an integer. And keep in mind that in many problems you start with quantities which need to be replaced by fractions in standard form [for example in trigonometry, problems are set in terms of $\cos(\theta)$ where $\theta$ is some angle].
But at the end of the day, it is just a convention. And while you think that $\frac{1}{\sqrt{2}}$ looks simpler, and you are right, the key with conventions is that they need to be consistent for the cases where you need recognition. The one which looks simpler is often relative...
The historical reason for rationalizing the denominator is that before calculators were invented, square roots had to be approximated by hand.
To approximate $\sqrt{n}$, where $n \in \mathbb{N}$, the ancient Babylonians used the following method:
Make an initial guess, $x_0$.
Let $$x_{k + 1} = \frac{x_k + \dfrac{n}{x_k}}{2}$$
If you use this method, which is equivalent to applying Newton's Method to the function $f(x) = x^2 - n$, to approximate the square root of $2$ by hand with $x_0 = 3/2$, you will see that while the sequence converges quickly, the calculations become onerous after a few steps. However, once an approximation was known, it was easy to calculate $$\frac{1}{\sqrt{2}}$$ quickly by rationalizing the denominator to obtain $$\frac{\sqrt{2}}{2}$$ then dividing the approximation by $2$.
I may have missed it, but there is an important reason that I think has been omitted from the other answers. (Ahaan Rungta mentioned it, but did not explain in detail.)
Recall how something like $\frac3{17}$ was calculated prior to around 1964:
$$\require{enclose} \begin{array}{rl} 17&\enclose{longdiv}{3.000\ldots} \end{array}$$
$$\begin{array}{rlll} & \ \ \ \,0.1\\ 17&\enclose{longdiv}{3.000\ldots} \\ & \ \ 1.7 \\ \hline & \ \ 1\ 3 \end{array}$$
$$\begin{array}{rlll} & \ \ \ \,0.17\\ 17&\enclose{longdiv}{3.000\ldots} \\ & \ \ 1.7 \\ \hline & \ \ 1\ 30 \\ & \ \ 1\ 19 \\ \hline & \ \ \ \ \ 11 \end{array}$$
And so on. The difficulty of the calculations depends only on the complexity of the divisor, which is 17. To extract a result with any required degree of precision one needs only continue the calculation until the required number of digits have been emitted. But the operations themselves are determined by the divisor.
Now let us take $\frac3{\sqrt2}$ as an example. To calculate this directly we need to evaluate:
$$1.4142\ldots \enclose{longdiv}{3.000\ldots} $$
which is quite onerous. Using an exact value for the divisor is impossible because of the way the algorithm works, so you must truncate the divisor. It's not clear how much error will be introduced by this truncation. And if you round off the divisor to $n$ digits of precision, you must perform many multiplications and subtractions of $n$-digit numbers.
In contrast, calculating $\frac{3\sqrt2}2$ is much easier. First calculate $3\times \sqrt2$ with a single multiplication, to obtain $4.242640\ldots$. (If you need more digits later you can easily produce them when you need them.)
Then perform the following division:
$$2 \enclose{longdiv}{4.242640\ldots} $$
which requires only trivial integer calculations throughout.