Yes, the maximal sum is the one of the equilateral triangle, that is $9R^2$.
Since Prove that in any triangle $ABC$, $\cos^2A+\cos^2B+\cos^2C\geq\frac{3}{4}$ then
$$
\sin^2 A+\sin^2 B+\sin^2 C=3-\cos^2 A-\cos^2 B-\cos^2 C\leq \frac{9}{4}
$$
where $A$, $B$ and $C$ are non negative numbers such that $A+B+C=\pi$. Hence, for any inscribed triangle, the sum of the squares of the sides is
$$(2R\sin A)^2+(2R\sin B)^2+(2R\sin C)^2\leq 9R^2.$$
OK, I think I was misremembering in my comment, I think the perimeter is easier to work with than the area. So you start with a circle of circumference $\pi$ (meaning a radius of $\frac{1}{2}$). Find the length of the side of the square (it will be $\frac{1}{\sqrt{2}}$), so the initial guess is $4\cdot\frac{1}{\sqrt{2}} = 2\sqrt{2} \approx 2.828427$:
Here is a concrete example where we can use the previous known chord (in this case $\frac{1}{\sqrt{2}}$) to find the next:
This gives $\pi \approx 3.0614674$. Now, here is the general case, where you know the previous chord, $s_n$, and then find the next (knowing that each time, you're bisecting the previous chord so the number of sides doubles). I think this formula is correct, the formula for $s_{n+1}$ given $s_n$ is definitely correct because I tested it, but not entirely certain about the perimeter formula:
Using the above, we can write:
$$
s_{n+1}^2 = \frac{1 - \sqrt{1 - s_n^2}}{2}
$$
So we can find:
\begin{align*}
P_1 \approx &\ 2.8284271247461903\\
P_2 \approx &\ 3.061467458920718\\
P_3 \approx &\ 3.121445152258053\\
P_4 \approx &\ 3.1365484905459406\\
P_5 \approx &\ 3.140331156954739\\
P_6 \approx &\ 3.141277250932757\\
P_7 \approx &\ 3.1415138011441455\\
P_8 \approx &\ 3.1415729403678827\\
P_9 \approx &\ 3.141587725279961\\
P_{11} \approx &\ 3.141591421504635\\
P_{12} \approx &\ 3.141592345611077\\
P_{13} \approx &\ 3.1415925765450043\\
P_{14} \approx &\ 3.1415926334632482\\
P_{15} \approx &\ 3.141592654807589\\
P_{16} \approx &\ 3.1415926453212153\\
P_{17} \approx &\ 3.1415926073757197\\
P_{18} \approx &\ 3.1415929109396727\\
P_{19} \approx &\ 3.141594125195191\\
P_{20} \approx &\ 3.1415965537048196\\
P_{21} \approx &\ 3.1415965537048196
\end{align*}
This gives five digits of precision: $\pi \approx 3.14159$.
This example was taught to me in my Scientific Computing class way back to display roundoff error in floating point calculations. You'll notice on the last two, we get the same result. That's because the floating point calculations of the computer have essentially hit their limit. The reason for this is because $s_n^2$ has become so small, that $1 - s_n^2$ doesn't generate a "new" number (it just keeps giving the same number that will generate $s_n^2$ again when subtracted from $1$. There is a trick to make this calculation better:
\begin{align*}
s_{n+1}^2 =&\ \frac{1 - \sqrt{1 - s_n^2}}{2}\cdot\frac{1 + \sqrt{1 - s_n^2}}{1 + \sqrt{1 - s_n^2}} \\
=&\ \frac{1}{2}\cdot\frac{1 - \left(1 - s_n^2\right)}{1 + \sqrt{1 - s_n^2}}\\
=&\ \frac{1}{2}\cdot\frac{s_n^2}{1 + \sqrt{1 - s_n^2}}
\end{align*}
This really does give better results:
\begin{align*}
P_1 \approx&\ 2.8284271247461903 \\
P_2 \approx&\ 3.0614674589207183\\
P_3 \approx&\ 3.1214451522580524\\
P_4 \approx&\ 3.1365484905459393\\
P_5 \approx&\ 3.140331156954753\\
P_6 \approx&\ 3.141277250932773\\
P_7 \approx&\ 3.1415138011443013\\
P_8 \approx&\ 3.1415729403670913\\
P_9 \approx&\ 3.1415877252771596\\
P_{10} \approx&\ 3.1415914215111997\\
P_{11} \approx&\ 3.1415923455701176\\
P_{12} \approx&\ 3.1415925765848725\\
P_{13} \approx&\ 3.1415926343385627\\
P_{14} \approx&\ 3.1415926487769856\\
P_{15} \approx&\ 3.141592652386591\\
P_{16} \approx&\ 3.141592653288993\\
P_{17} \approx&\ 3.141592653514593\\
P_{18} \approx&\ 3.141592653570993\\
P_{19} \approx&\ 3.1415926535850933\\
P_{20} \approx&\ 3.141592653588618\\
P_{21} \approx&\ 3.1415926535894996\\
P_{22} \approx&\ 3.1415926535897203\\
P_{23} \approx&\ 3.1415926535897754\\
P_{24} \approx&\ 3.141592653589789\\
P_{25} \approx&\ 3.1415926535897927\\
P_{26} \approx&\ 3.1415926535897936\\
P_{27} \approx&\ 3.1415926535897936\\
\end{align*}
By simply changing the computation, not the algorithm!, we now get twelve digits of precision! $\pi \approx 3.141592653589$--all because of floating point roundoff error.
Best Answer
This is a question about linguistics and psychology and teaching, not really about mathematics.
We have special words for things we refer to often. Circles come up way more often than ellipses so it's convenient (and clear historically) that they have their own word. "Square" is much nicer than "equilateral rectangle" and requires a lot less cognitive processing.
I spend a fair amount of time in K-5 classrooms, so I've some experience with the questions you raise.
Yes, kids in elementary school are confused by the fact that a square is a rectangle. So are some elementary school teachers. That's a problem with trying to impose correct formal mathematics on informal everyday speech. It happens a lot - this is an instance (in a way) of whether "or" means "and or" or "or but not and". In mathematics it's always the former. In daily life, sometimes one sometimes the other.
One problem I have with elementary school "geometry" is its focus on categorizing and naming things and its paucity of theorems - or at least observations of properties. I wish kids were taught to notice that the diagonals of a parallelogram bisect each other, or that the medians of a triangle meet at a point, way before they encounter proofs.
And yes, this should be migrated to math education SE.