negative/inverse triangular inequality, asymptotic analysis

Question

For $r$ sufficiently large and $|z|\ge r$ we have $$|z^2-4z+3|\ge||z|^2-|4z-3||\ge|z|^2-4|z|-3\ge\frac{|z|^2}{2}\ge{r^2\over2}$$

How do we get the second to last inequality?

Is it still true if we replace $2$ by any $\alpha\in \Bbb R_{>1}$?

Answer 1

For $|z| \ge r > 0$: $$ |z|^2-4|z|-3 = |z|^2 \left( 1 - \frac{4}{|z|} -\frac{3}{|z|^2} \right) \ge |z|^2 \left( 1 - \frac{4}{r} -\frac{3}{r^2} \right) $$ For $r > 0$ sufficiently large the second factor is $\ge \frac 12$.

Answer 2

Parabolas grow faster than lines. That is, $1/2x^2>ax+b$ for $x$ large enough for all $a,b$. So if $|z|$ is sufficiently large, $1/2|z|^2>4|z|+3$.

Answer 3

Concerning your second question about replacing $2$ by $\alpha > 1$:

Yes, the claim is still true for sufficiently large $r$.

Let $\beta = 1-\frac{1}{\alpha}$, hence $\beta \in (0,1)$: \begin{eqnarray*} |z|^2-4|z|-3 &\ge & \frac{|z|^2}{\alpha }\\ &\Leftrightarrow & \\ \beta|z|^2 -4|z| &\ge & 3\\ &\Leftrightarrow & \\ |z|\left( \beta|z| -4\right) &\ge & 3\\ \end{eqnarray*} The last inequality is surely satisfied for $\boxed{|z| > \max\left( \frac{7}{\beta},3 \right)}$.