Find an maxima such that $\sqrt a+\sqrt b$, when $27a^2+b^2=27$ (In high-school level!)

Question

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.

Answer 1

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.)

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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 $$1=\frac{27x^3}{(27-27x^4)^{3/4}} \implies x=\frac{\sqrt 2}2.$$

Answer 2

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 $(x, y) = (\frac{1}{\sqrt{2}}, \frac{3}{\sqrt{2}})$.

Answer 3

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.

Answer 4

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.