How to prove that $1/|z^4-4z^2+3|\le 1/3$ if $z$ is a complex number with $|z|=2 $?

Question

Show $$\left\lvert \frac{1}{z^4-4z^2+3} \right\rvert \leq \frac{1}{3},\, \text{ if } |z|=2.$$

I am sure it is pretty easy and I am overlooking something.

So this is equivalent to $$3 \leq \left\lvert z^2-1\right\rvert\cdot\left\lvert z^2-3\right\rvert.$$ I know $z^2$ is on the circle of radius 4 centered at the origin in $\mathbb{C}$. So I am now thinking of the distance from $z^2$ to $1$ and the distance from $z^2$ to $3$ to help show the bound is true. I am trying to find the minimum distances of $z^2$ to $1$ for example, and the way I have been going about this is by explicitly looking at the distance and making a function $f(x,y)$ and trying to compute its minimum, but I don't know how to really do this without breaking into cases of $x\geq 0$ and $x <0$ if $z=x+iy$ so I am wondering if this is overkill and if there is another simpler way.

Answer 1

Note that, on $|z|=2$, $$|z^4-4z^2+3|=|(z^2-1)(z^2-3)|\ge(|z|^2-1)(|z|^2-3)=3$$ and hence $$\frac{1}{|z^4-4z^2+3|}\le\frac13.$$

Answer 2

Hint: $|z^2 - 1| = |4\mathrm e^{i\theta}-1|$ is a circle of radius $4$ centred at $-1$. It is therefore minimised when $\theta = 0$. Something similar will work for the other factor.

Answer 3

Your inequality is equivalent to: $$\forall z\in S^1,\qquad |16z^4-16z^2+3|\geq 3,$$ or to: $$\forall \theta\in[0,2\pi],\quad \left(16\cos(4\theta)-16\cos(2\theta)+3\right)^2+\left(16\sin(4\theta)-16\sin(2\theta)\right)^2\geq 9,$$ that is equivalent to: $$ 512+9-512\cos(2\theta)+96\left(\cos(4\theta)-\cos(2\theta)\right)\geq 9$$ or to: $$ 16-19\cos(2\theta)+3\cos(4\theta)\geq 0$$ or to: $$ \sin^2\theta\left(13-6\cos(2\theta)\right)\geq 0$$ that is trivial.