Find maximum without calculus 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?

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 $\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3} = 1$.
