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?

There's always the classic, but perhaps computationally icky derivative approach.$$f(x)=\dfrac{\sqrt{3}}{4}x^2+\dfrac{\sqrt{(9-x^2)(x^2-1)}}{4}$$
$$f'(x)= \frac{\sqrt3}2 x + \frac{20 x - 4 x^3}{
 8 \sqrt{(9 - x^2) (-1 + x^2)}} = \frac 12 x \, \Big(\sqrt{3} + \frac{5 - x^2}{\sqrt{- x^4 - 10 x^2 - 9 }}\Big)$$
Let $f'(x) = 0$ and solve to find the local extrema:
Clearly $x=0$ is a solution, but doesn't exist in the domain of $f$, which is $[−3,−1]∪[1,3]$.
So we find solutions to $$\sqrt{3} + \frac{5 - x^2}{\sqrt{- x^4 - 10 x^2 - 9 }} = 0$$ or equivalently $$ \sqrt3 \sqrt{-x^4+10 x^2-9} = 5 - x^2 $$ which comes to
$$3 (-x^4+10 x^2-9)=(5-x^2)^2$$
$$-3 x^4+30 x^2-27=x^4-10 x^2+25$$
$$-4 x^4+40 x^2-52=0$$
Complete the square to get
$$(x^2-5)^2=12$$ so
$$x = \pm \sqrt{5 \pm 2 \sqrt 3}$$
Checking for extraneous solutions reveals that the only values of $x$ satisfying $f'(x) = 0$ are $$x = \pm \sqrt{5 +2 \sqrt 3}$$
Now plug in these values to get
$$f_{\text{max}} = f(\pm \sqrt{5 +2 \sqrt 3}) = \frac{8 + 5 \sqrt3}4$$ which is in line with MyGlasses's observation that $f_{\text{max}}$ must be no more than $\dfrac{\sqrt{3}}{4}9+\dfrac{8}{4}$.
