Maximise $(x+1)\sqrt{1-x^2}$ without calculus Problem
Maximise $f:[-1,1]\rightarrow \mathbb{R}$, with $f(x)=(1+x)\sqrt{1-x^2}$
With calculus, this problem would be easily solved by setting $f'(x)=0$ and obtaining $x=\frac{1}{2}$, then checking that $f''(\frac{1}{2})<0$ to obtain the final answer of $f(\frac{1}{2})=\frac{3\sqrt{3}}{4}$
The motivation behind this function comes from maximising the area of an inscribed triangle in the unit circle, for anyone that is curious.
My Attempt
$$f(x)=(1+x)\sqrt{1-x^2}=\sqrt{(1-x^2)(1+x)^2}=\sqrt 3 \sqrt{(1-x^2)\frac{(1+x)^2}{3}}$$
By the AM-GM Inequality, $\sqrt{ab}\leq \frac{a+b}{2}$, with equality iff $a=b$
This means that
$$\sqrt 3 \sqrt{ab} \leq \frac{\sqrt 3}{2}(a+b)$$
Substituting $a=1-x^2, b=\frac{(1+x)^2}{3}$,
$$f(x)=\sqrt 3 \sqrt{(1-x^2)\frac{(1+x)^2}{3}} \leq \frac{\sqrt 3}{2} \left((1-x^2)+\frac{(1+x)^2}{3}\right)$$
$$=\frac{\sqrt 3}{2} \left(\frac{4}{3} -\frac{2}{3} x^2 + \frac{2}{3} x\right)$$
$$=-\frac{\sqrt 3}{2}\frac{2}{3}(x^2-x-2)$$
$$=-\frac{\sqrt 3}{3}\left(\left(x-\frac{1}{2}\right)^2-\frac{9}{4}\right)$$
$$\leq -\frac{\sqrt 3}{3}\left(-\frac{9}{4}\right)=\frac{3\sqrt 3}{4}$$
Both inequalities have equality when $x=\frac{1}{2}$
Hence, $f(x)$ is maximum at $\frac{3\sqrt 3}{4}$ when $x=\frac{1}{2}$
However, this solution is (rather obviously I think) heavily reverse-engineered, with the two inequalities carefully manipulated to give identical equality conditions of $x=\frac{1}{2}$. Is there some better or more "natural" way to find the minimum point, perhaps with better uses of AM-GM or other inequalities like Jensen's inequality? 

You can solve the problem using geometry. 
$(x+1)\sqrt{1-x^2}$ is the area of triangle 
$$(-1, 0), (x, \sqrt{1-x^2}), (x, -\sqrt{1-x^2}).$$
This unit inscribed triangle has maximal area if and only if it is a equilateral triangle.
