Find maximum without calculus

Question

Let $f:(0,1]\rightarrow\mathbb{R}$ with $f(x)=2x(1+\sqrt{1-x^2})$. Is it possible to find the maximum of this function without calculus? Possibility through some series of inequalities?

Answer 1

$(f(x))^2 = 4x^2\left(1+\sqrt{1-x^2}\right)^2$. Let $y = \sqrt{1-x^2} \Rightarrow x^2 = 1 - y^2 \Rightarrow (f(x))^2 = 4(1-y^2)(1+y)^2 = 4(1-y)(1+y)^3$.

Apply AM-GM inequality we have:

$2 = (1-y) + \dfrac{1+y}{3} + \dfrac{1+y}{3} + \dfrac{1+y}{3} \geq 4\sqrt[4]{\dfrac{(1-y)(1+y)^3}{27}} \Rightarrow \dfrac{16\times 27}{4^4} \geq (1-y)(1+y)^3 \Rightarrow \dfrac{27}{16} \geq (1-y)(1+y)^3 \Rightarrow \dfrac{27}{4} \geq 4(1-y)(1+y)^3 = (f(x))^2 \Rightarrow f(x) \leq \sqrt{\dfrac{27}{4}} = \dfrac{3\sqrt{3}}{2} \Rightarrow f_{\text{max}} = \dfrac{3\sqrt{3}}{2}$.

$=$ occurs when $1-y = \dfrac{1+y}{3} \Rightarrow y = \dfrac{1}{2} \Rightarrow \sqrt{1-x^2} = \dfrac{1}{2} \Rightarrow x = \dfrac{\sqrt{3}}{2}$.

Answer 2

I'll admit its essentially same as what ah-huh-moment did here, but here's another approach:

Make the substitution $x = \sin \theta$, for $\displaystyle \theta \in \left(0,\pi/2\right)$,

then, $\displaystyle f(x) = x + 2x\sqrt{1-x^2} = 2\sin \theta + \sin (\pi - 2\theta) \le 3\sin \frac{2\theta + \pi - 2\theta}{3} = \frac{3\sqrt{3}}{2}$

from Jensen's Inequality.

Answer 3

Putting in some more details to make the answer better: \begin{align*} f(x) &= 2x + 2x\sqrt{1-x^2}\\ &= 2\sin(\theta) + 2\sin(\theta)\sqrt{\smash[b]{1-\sin(\theta)^2}} &\text{via \( x = \sin(\theta)\), $\theta \in (0,\frac{\pi}{2}]$}\\ &= 2\sin(\theta)+2\sin(\theta)\cos(\theta)\\ &= 2\sin(\theta) + \sin(2\theta)\\ &= 3\left(\frac{1}{3}\sin(\theta)+\frac{1}{3}\sin(\theta)+\frac{1}{3}\sin(2\theta)\right)\\ &= 3\left(\frac{1}{3}\sin(\theta)+\frac{1}{3}\sin(\theta)+\frac{1}{3}\sin(\pi-2\theta)\right)\\ &\leq 3\sin\left(\frac{1}{3}\theta+\frac{1}{3}\theta+\frac{1}{3}(\pi-2\theta)\right) &\text{by Jensen's Inequality}\\ &= 3\sin\left(\frac{\pi}{3}\right)\\ &= \frac{3\sqrt{3}}{2} \end{align*} Note that $\sin(\cdot)$ is concave down (convex) when $\theta$ is in $\left(0, \dfrac{\pi}{2}\right)$ and $\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3} = 1$.