Find an maxima such that $\sqrt a+\sqrt b$, when $27a^2+b^2=27$ (In high-school level!) How to get a maximum value of $\sqrt a+\sqrt b$, when $27a^2+b^2=27$?

I know that using Lagrange multiplier is easiest way in college-level math, but I believe there is a elementary and elegant way to prove that in high-school level.

I made $a=\cos\theta$ and $b=3\sqrt3\sin\theta$, but in that case, the process of finding the maximum value becomes too messy.

I need your help.

Let $x = \sqrt{a}$ and $y = \sqrt{b}$, so we are aiming
\begin{align*}
\max_{\substack{x, y \ge 0 \\ 27x^4 + y^4 = 27}} (x + y)
\end{align*}
We can find the max via two applications of Cauchy-Schwarz:
\begin{align*}
x + y = \left\langle \left(x, \frac{y}{\sqrt{3}}\right), (1, \sqrt{3}) \right\rangle \le \sqrt{x^2 + \frac{y^2}{3}}\sqrt{4} \\
x^2 + \frac{y^2}{3} = \left\langle \left(x^2, \frac{y^2}{3\sqrt{3}}\right), (1, \sqrt{3}) \right\rangle \le \sqrt{x^4 + \frac{y^4}{27}}\sqrt{4} = \sqrt{4}
\end{align*}
In order words
\begin{align*}
x + y \le \sqrt{\sqrt{4}} \sqrt{4} = 2\sqrt{2}
\end{align*}
To show this bound is attainable, plug in $(x, y) = (\frac{1}{\sqrt{2}}, \frac{3}{\sqrt{2}})$.
