<s>[INST] 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?
 [/INST] We can simplify the problem by finding the minimun of
$$ g(t) = \frac{\sqrt{3}}{4}t + \frac{\sqrt{(9-t)(t-1)}}{4} $$
with $t = x^2, \ t \in [1,9]$
A straightforward approach is just to take the first derivative
$$ g'(t) = \frac{\sqrt{3}}{4} + \frac{5-t}{4\sqrt{-9+10t-t^2}} $$
Solving $g'(t)=0$ leads to
$$ \sqrt{16 - (t-5)^2} = \frac{t-5}{\sqrt{3}} \quad (t > 5) $$
which gives $t = 5 + 2\sqrt{3}$
Thus, the maximum is 
$$g(5+2\sqrt{3}) = \frac{\sqrt{3}}{4}(5+2\sqrt{3}) + \frac{1}{2} = 2 + \frac{5\sqrt{3}}{4}$$
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