<s>[INST] 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.
 [/INST] 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
$$1=\frac{27x^3}{(27-27x^4)^{3/4}} \implies x=\frac{\sqrt 2}2.$$
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