Find maximum without calculus

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

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.