Find this maximum of the $\frac{\sqrt{3}}{4}x^2+\frac{\sqrt{(9-x^2)(x^2-1)}}{4}$ Let $x\in \mathbb{R}$, find the function maximum of the value
$$f(x)=\dfrac{\sqrt{3}}{4}x^2+\dfrac{\sqrt{(9-x^2)(x^2-1)}}{4}$$
my attemp
$$x^2=5+4\sin{t},t\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$
then
$$f=\dfrac{5\sqrt{3}}{4}+2\sin{\left(t+\frac{\pi}{6}\right)}\le 2+\dfrac{5}{4}\sqrt{3}$$
My Question:this function have other methods to find this maximum? such as AM-GM,Cauchy-Schwarz inequality and so on?

Consider the function and its first derivative$$f(x)=\frac{\sqrt{3}}{4}x^2+\frac{\sqrt{(9-x^2)(x^2-1)}}{4}$$
$$f'(x)=\frac{1}{2} x \left(\frac{5-x^2}{\sqrt{-x^4+10 x^2-9}}+\sqrt{3}\right)$$ The first derivative cancels for $x=0$  which has to be excluded.
Now, let $x^2=y$ and solve
$$\frac{5-y}{\sqrt{-y^2+10 y-9}}+\sqrt{3}=0\implies 5-y=-\sqrt{3}\sqrt{-y^2+10 y-9}$$ and square to get $$4 y^2-40 y+52=0 \implies y=5\pm2 \sqrt{3}$$ Only $y=5+2 \sqrt{3}$ must be kept because of the real domain.
$$f(\sqrt{5+2 \sqrt{3}})=2+\frac{5 \sqrt{3}}{4}$$
