The sum of the floor function counts the integer points lying on or below a branch of the rectangular hyperbola $xy=n$ (for $x,y\gt 0$). Sketch this and add the square with corners $(0,0),(0,\sqrt n), (\sqrt n, 0), (\sqrt n, \sqrt n)$ (the last of these points lies on the hyperbola).
$k^2$ counts the points in the square, and the tails are symmetrical about $x=y$. The sum taken once counts the points in the tail, plus the points in the square (which are therefore counted twice when the sum is multiplied by $2$).
You should be able to turn these observations into a formal proof.
This is now the fourth time (at least) that a different form of question about this sum has been posed recently. The sum is equivalent to the sum of the divisor function $d(n)$, which counts the number of divisors of $n$.
Further note
If $(a,b)$ is below the hyperbola xy=n, so is $(b,a)$. Look at the sum on the right-hand side, taken once. It counts the number of points $(x,y)$ on the line $x=1$ with $0\lt xy \le n$ i.e. $\left\lfloor \frac n1 \right\rfloor$, the points with $x=2$ i.e. $\left\lfloor \frac n2 \right\rfloor$, up to the points with $x=k$ ie $\left\lfloor \frac nk \right\rfloor$.
Note that the points $(a,b)$ with $a,b \le k$ are all counted in this sum. No points with $a \gt k$ are counted.
The sum also counts the points in a similar way for $y=1, 2 \dots k$. This also has the points $(a,b)$ with $a,b\le k$ - we note there are $k^2$ of these, but has no points with $b\gt k$.
If we count the sum twice, we have counted all the points under the hyperbola, but there are $k^2$ points we have counted twice, so we subtract $k^2$ to avoid double-counting.
For other questions see this, and this.
In step $5$ you are going to multiply the digit $d$ you put on top by $20+d$ because you write $d$ to the right of the $2$ and multiply. To get a positive remainder you need $d(20+d) \le 174$. Here if $d=6, d(20+d)=156$, while if $d=7, d(20+d)=189,$ which is too large.
The idea of the algorithm is based on $(a+b)^2=a^2+2ab+b^2$. $a$ represents the digits you have already found, and you have subtracted $a^2$ from the number you are taking the square root of already. In your example, the first $a$ is $100$ and we subtracted $100^2$ in step $3$. Writing twice the current set of digits (here $2$) is the $2a$ part of $2ab+b^2$. When you write the next digit in both places and multiply, you get $2ab+b^2$. When you subtract that you have subtracted $a^2+2ab+b^2$ from the original number, here leaving $27493-25600=1893$. Our new $a$ is $160$ and we need to find $b$ such that $2ab+b^2 \lt 1893$, which turns out to be $5$.
Best Answer
This is from the comment of André Nicolas, which I will just attempt to clarify. Assume $x,y\ge0$. We want $${[\sqrt x + \sqrt y]\over 2 } \le \sqrt{(x+y)\over 2}$$
We proceed with inequalities equivalent to the first and to each other:
$${[\sqrt x + \sqrt y]\over 2 } \le {\sqrt{(x+y)}\over \sqrt 2}$$ $${\sqrt x + \sqrt y } \le {2\sqrt{(x+y)}\over \sqrt 2}$$ $${\sqrt x + \sqrt y } \le {\sqrt2\sqrt{(x+y)}}$$ squaring $$x+2\sqrt{xy}+y\le 2x+2y$$ $$x+y-2\sqrt{xy}\ge0$$ $${{[\sqrt x - \sqrt y]}}^2\ge0$$ which we know to be true. Thus, its equivalent inequality, the one we are trying to prove is also true.