Minimize the ratio involving the ellipse 
Let $P$ be any point on the curve $\dfrac{x^2}{4}+\dfrac{y^2}{3}=1$,
and $A,B$ be two fixed points $\left(\dfrac{1}{2},0\right)$ and
$(1,1)$ respectively. Find the minimum value of
$\dfrac{|PA|^2}{|PB|}$.

Assume $x=2\cos\theta,y=\sqrt{3}\sin\theta$, then
$$\dfrac{|PA|^2}{|PB|}=\frac{\left(2\cos\theta-\frac{1}{2}\right)^2+(\sqrt{3}\sin\theta-0)^2}{\sqrt{\left(2\cos\theta-1\right)^2+(\sqrt{3}\sin\theta-1)^2}},$$
but which is not so easy to tackle.
WolframeAlpha gives the result $1$.

Doing what @Paul Sinclair already commented and using multiple angle formulae
$$\dfrac{|PA|^4}{|PB|^2}=\frac{-256 \cos (\theta )+92 \cos (2 \theta )-16 \cos (3 \theta )+2 \cos
   (4 \theta )+259}{8 \left(-4 \sqrt{3} \sin (\theta )-8 \cos (\theta
   )+\cos (2 \theta )+11\right)}$$ Computing the derivative, its numerator is "just"
$$(15-8 \cos (\theta )+2 \cos (2 \theta )) \times$$
$$\left(24 \sqrt{3}-40 \sin (\theta )+45 \sin (2 \theta )-12 \sin (3 \theta )+\sin
   (4 \theta )-40 \sqrt{3} \cos (\theta )-8 \sqrt{3} \cos (2 \theta )+6
   \sqrt{3} \cos (3 \theta )\right)$$ The first factor does not cancel and with all these $\sqrt{3} $ in the remaining term, if the solution is simple, it is a multiple of $\frac \pi 3$ and, just trying, one zero of the derivative is $\theta=\frac {5\pi} 3$ and then the result.
