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.
 A: Here's an entirely elementary solution, although it may not be very satisfying. We have, using the power-mean inequality,
\begin{align*}
\frac{\sqrt a + \sqrt b}4
&=\frac{\sqrt a + \sqrt{\frac b9} + \sqrt{\frac b9} + \sqrt{\frac b9}}4\\
&\leq \left(\frac{(\sqrt a)^4+\left(\sqrt{\frac b9}\right)^4+\left(\sqrt{\frac b9}\right)^4+\left(\sqrt{\frac b9}\right)^4}{4}\right)^{1/4}\\
&=\left(\frac{a^2+\frac{b^2}{81}+\frac{b^2}{81}+\frac{b^2}{81}}{4}\right)^{1/4}\\
&=\left(\frac{\frac{27a^2+b^2}{27}}{4}\right)^{1/4}=\frac1{4^{1/4}}=\frac1{\sqrt 2}.
\end{align*}
As a result, $\sqrt a+\sqrt b\leq 2\sqrt 2$. Equality is reached at $a=1/2$ and $b=9/2$, as can be traced through the application of the power-mean inequality.
The power-mean inequality is the statement that
$$\left(\frac{\sum_{i=1}^n a_i^p}{n}\right)^{1/p}$$
is an increasing function in $p$. This can be proven for the necessary exponents ($1$ and $4$) in four variables by first showing it for two variables:
$$\frac{x+y}{2}\leq \left(\frac{x^4+y^4}2\right)^{1/4}$$
since
$$\frac{x^4+y^4}2-\left(\frac{x+y}2\right)^4=\frac{(x+y)^4(7x^2+10xy+7y^2)}{16},$$
and then using this to show
$$\frac{w+x+y+z}{4}\leq \frac{\left(\frac{w^4+x^4}2\right)^{1/4}+\left(\frac{y^4+z^4}2\right)^{1/4}}2\leq \left(\frac{w^4+x^4+y^4+z^4}{4}\right)^{1/4}.$$
(These techniques are similar to the proof of the AM-GM inequality.)

It's probably easier to simply use differential calculus: letting $x=\sqrt a$, we wish to maximize
$$f(x) = x+(27-27x^4)^{1/4}.$$
We compute
$$f'(x)=1-\frac{27x^3}{(27-27x^4)^{3/4}}; f''(x)=-\frac{3^{7/4}x^2}{(1-x^4)^{7/4}}<0,$$
so the function has a unique maximum where
$$1=\frac{27x^3}{(27-27x^4)^{3/4}} \implies x=\frac{\sqrt 2}2.$$
A: 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 $(x, y) = (\frac{1}{\sqrt{2}}, \frac{3}{\sqrt{2}})$.
A: You can homogeneise a bit the coefficients by setting $\begin{cases}a=A^2\\b=9B^2\end{cases}$ to get to $\begin{cases}A^4+3B^4=1\\\min(A+3B)\end{cases}$
In fact substituting $A=(1-3B^4)^{\frac 14}$ does not lead to such a mess for the derivative :
$\dfrac{\partial(A+3B)}{\partial B}=3-3\dfrac{B^3}{(1-3B^4)^{\frac 34}}=0\iff (1-3B^4)^{\frac 34}=B^3\implies B^{12}=(1-3B^4)^3$
This simplifies to $$28B^{12}-27B^8+9B^4-1=\underbrace{(2B^2+1)}_\text{>0}(2B^2-1)\underbrace{(7B^8-5B^4+1)}_{>0}$$
Note: the last one behaves as $7x^2-5x+1$, which is a quadratic with no real roots.
So in the end, only $B^2=\frac 12$ which leads to $A^2=\frac 12$ too are critical values.
A: Alternately, using Hölder's Inequality,
$$\sqrt{a}+\sqrt{b} \leqslant \left(27a^2+b^2 \right)^{1/4} \left(\tfrac13+1 \right)^{3/4} =2\sqrt2 $$
Equality is possible when $27a^2:b^2=\frac13:1 \implies (a, b) = (\frac12, \frac92)$, so this gives a maximum.
