The question is to find the upper bound of
$$\left|\frac{1}{z^4 + 3z^2 + 2}\right| ,\; where\;|z| = 2,\;z\; \epsilon\; \mathbb{C}$$
My Attempt:
To find its upper bound, finding the lower bound of denominator is sufficient.
Using the triangle inequality $|z_1 \pm z_2| \geqslant ||z_1| - |z_2||$,
$$|z^4 + 3z^2 + 2| \geqslant ||z^4| - |3z^2 + 2||$$
$$ \implies |z^4 + 3z^2 + 2| \geqslant |16 - |3z^2 + 2||$$
Now, the greatest value of $|3z^2 + 2|$ is required so that the difference is the smallest.
Using the triangle inequality $|z_1 + z_2| \leqslant |z_1| + |z_2|$,
$$ \implies |z^4 + 3z^2 + 2| \geqslant |16 - (|3z^2| +||2|)| = 2$$
So, this gives the upper bound to be $\frac{1}{2}$.
But, the answer is $\frac{1}{6}$, which I got as well when I solved it geometrically. I have got a hunch that my answer is wrong because I have used two triangle inequalities i.e. using one equality after another leads to an approximation.
Put $z^{2}=c$. The problem reduces to one of finding the minimum of $|c^{2}+3c+2|$ for $|c|=4$. Note that $|c^{2}+3c+2|=|(c+2)(c+1)| \geq (4-1)(4-2)=6$.
Hence $\frac 1 6$ is an upper bound for the given function. This bound is attained when $z=2i$.