A weird inequality Let $f(x) = 2x(\frac{\sqrt{3}}{2} -x)^2 - n(x^2+\frac{1}{4})^{\frac{3}{2}}$. Find the least positive value of $n>1$ such that $f(c)\neq 0, \forall c \in [0,\frac{\sqrt{3}}{2}]$ 
All my approaches have been to waste till now as as I have found not so useful bounds using second derivative tests.
Firstly I tried differentiating $f(x)$ to get $f'(x)=6x^2-4\sqrt{3}x+\dfrac{3}{2}-\dfrac{3}{2}nx(4x^2+1)^{\frac{1}{2}}.$ Now observing that $f'(0)=\dfrac{3}{2}$ and $f'(\dfrac{\sqrt{3}}{2})=-\dfrac{3\sqrt{3}}{2}n$ thus giving me a basic idea of what the graph might look like. I figured that there must be a local maxima in $[0,\dfrac{\sqrt{3}}{2}]$ and if the value of $f(x)$ at local maxima will tend to zero, we will have a nice bound on $n$ then. That was all of my progress and after this I am stuck as I am not able to solve at this condition.
Another approach I tried was using inequalities. Noticing $f(0)$ is negative, we can convert the above problem to the inequality $n(x^2+\frac{1}{4})^{\frac{3}{2}} >2x(\frac{\sqrt{3}}{2} -x)^2 \ \ \forall x \in [0,\frac{\sqrt{3}}{2}]$ but again this was pretty much another dead end.
 A: From your last line we get $f(x)\neq 0$ for all $x \in [0, \frac{\sqrt{3}}{2}]$ if and only if $$n>n^* \equiv\max_{x \in \left[0, \frac{\sqrt{3}}{2}\right]}\left\{ \frac{2x(\frac{\sqrt{3}}{2}-x)^2}{(x^2+1/4)^{3/2}}\right\}$$
So there is no "least positive value of $n$" that works, but $n^*$ is the infimum over all positive $n$ that work.  We want to calculate $n^*$.
Define
$a(x) =  2x(\frac{\sqrt{3}}{2}-x)^2$ and $b(x) =(x^2+1/4)^{3/2}$ and $g(x)=\frac{a(x)}{b(x)}$. We want to maximize $g$ over the interval.
Note that $g$ is differentiable. Also, $g$ is zero on the endpoints of the interval and positive in between, so its maximum occurs at a point inside the interval where the derivative is zero.   Then
$$ g'(x)=0 \implies b(x)a'(x)=a(x)b'(x)$$
Simplifying this expression leads to a cubic equation in $x$, but the $x^3$ terms cancel so it is really a quadratic in $x$. The only solution in the appropriate interval is
$$ x^* = \frac{\sqrt{11}-\sqrt{3}}{8} \approx 0.198072$$ so that $n^*=g(x^*)\approx 1.13625$. Now if you want to choose $n$ as an integer you can choose $n=2$.
