Maximizing $x^2y$ given $x^2+y^2=100$, without using the AM-GM inequality and calculus tools Problem says:

Let $x^2+y^2=100$, where $x,y>0$. For which ratio of $x$ to $y$, the value of $x^2y$ will be maximum?

I know these possible tools:

*

*AM-GM inequality


*Calculus tools

Here, I want to escape from all of the tools I mentioned above.
I will try to explain my attempts in the simplest sentences. (my english is not enough, unfortunately). I will not prove any strong theorem and also I'm not sure what I'm doing exactly matches the math, rigorously.

Solution I made:
First, it is not necessary to make these substitutions.  I'm just doing this to work with smaller numbers.
Let, $x=10m, ~y=10n$,
where $0<m<1,~ 0<n<1$, then we have
$$ x^2+y^2=100 \iff m^2+n^2=1$$
$$x^2y=1000m^2n$$
This means,
$$\max\left\{x^2y\right\}=10^3\max\left\{m^2n\right\}$$
$$m^2n=n(1-n^2)=n-n^3$$
Then suppose that,
$$\begin{align}\max\left\{n-n^3 \mid 0<n<1\right\}&=a, a>0&\end{align}$$
This implies
$$n-n^3-a≤0,~ \forall n\in\mathbb (0,1)$$
$$n^3-n+a≥0,~\forall n\in\mathbb (0,1)$$
Then, we observe that
$$\begin{align}n^3-n+a≥0, \forall n\in (0,1) ~ \text{and} ~ \forall n≥1\end{align}$$
This follows
$$ n^3-n+a≥0, ~ \forall n>0.$$

Using the last conclusion, I assume  that there exist $u,v>0$, such that

$$n^3-n+a=(n-u)^2(n+v)≥0.$$
If $n>0$, then the equality occurs, if and only if
$$n=u>0$$


Based on these, we have:
$$\begin{align}n^3-n+a= (n-u)^2(n+v)≥0 \end{align}$$
$$\begin{align}n^3-n+a = & n^3 - n^2(2u-v)+ n(u^2 - 2 u v ) + u^2v & \end{align}$$
$$\begin{align} \begin{cases} 2u-v=0 \\ u^2-2uv=-1 \\u^2v=a \\u,v>0 \end{cases} &\implies \begin{cases} v=2u \\ u^2-4u^2=-1 \\ 2u^3=a \\ u,v>0 \end{cases}\\
&\implies \begin{cases}  u=\frac{\sqrt 3}{3} \\ v=\frac{2\sqrt 3}{3}\\ a=2\left(\frac{\sqrt 3}{3} \right)^3=\frac{2\sqrt 3}{9} \end{cases} \end{align}$$
$$\begin{align}n^3-n+\frac{2\sqrt 3}{9} &=\left(n-\frac{\sqrt 3}{3} \right)^2\left(n+\frac{2\sqrt 3}{3}\right)≥0.&\end{align}$$
As a result, we deduce that
$$\begin{align}n-n^3-\frac{2\sqrt 3}{9} &=-\left(n-\frac{\sqrt 3}{3} \right)^2\left(n+\frac{2\sqrt 3}{3}\right)≤0, &\forall n\in (0,1).&\end{align}$$
$$\begin{align}\max\left\{n-n^3 \mid 0<n<1\right\}&=\frac{2\sqrt 3}{9}, ~ \text{at }~ n=\frac{\sqrt 3}{3}&\end{align}$$
Finally, we obtain
$$m=\sqrt{1-n^2}=\sqrt{1-\frac 13}=\frac{\sqrt 6}{3}$$
$$\frac xy=\frac mn=\sqrt 2.$$

Question:

*

*How much of the things I've done here are correct?


An alternative approach without calculus. Considering that $x^2y = \lambda$ and $x^2+y^2=10^2$ are analytic, at it's maximum, $x^2y$ should be tangent to $x^2+y^2=10^2$ then
$$
\frac{\lambda}{y}+y^2-10^2=0
$$
so, at tangency $y^3-10^2y+\lambda = (y-r_1)^2(y-r_2)$
Equating coefficients we have
$$
\cases{
r_1^2r_2 +\lambda = 0\\
r_1^2+2r_1r_2+10^2=0\\
2r_1+r_2 = 0
}
$$
now solving, we have
$$
\cases{
r_1 = \pm\frac{10}{\sqrt{3}}\\
r_2 = \mp\frac{20}{\sqrt{3}}\\
\lambda = \frac{2000}{3\sqrt{3}}
}
$$
so the maximum is $\frac{2000}{3\sqrt{3}}$ and choosing $y^* = \frac{10}{\sqrt{3}}$ we have $x^* = 10\sqrt{\frac 23}$
Attached a plot showing in red $x^2+y^2=10^2$ and in black the level curves of $x^2y$

