Find an upper bound for a modulus of a complex number

Question

I need to find an upper bound for the modulus $ |3z^2+2z+1| $ if $ |z| \leq 1$

My solution is:
$$ |3z^2+2z+1| = |(3z-1)(z+1)+2|$$

Using the next 2 equations from triangle inequality $$(1).|z_{1}+z_{2}| \leq |z_{1}|+|z_{2}| $$ $$(2).|z_{1}-z_{2}| \geq ||z_{1}|-|z_{2}|| $$

Using eq.(1) I got $$|(3z-1)(z+1)+2|\leq|(3z-1)(z+1)|+|2| ,\; \; which \; is \; equal\; to\;\; |3z-1||z+1|+2$$

Using eq.(2) in $|3z-1|$ I got $$|3z-1|\geq||3z|-|1||=|3|z|-1|$$

Using eq.(1) in $|z+1|$ I got $$|z+1|\leq|z|+|1|=|z|+1$$

Finally, substituting I got $$(|3|z|-1|)(|z|+1)+2$$

Using $|z|\leq1$ $$(|3(1)-1|)((1)+1)+2 = 6$$

The book says the result is 6, however I'm not sure about my procedure, so I want to know:

Is my procedure wrong, and I'm getting the correct answer by coincidence? If it's not, Is there a better way to solve it?

Answer 1

$|3z^2+2z+1| \leq |3z^2| + |2z| + |1| =3|z|^2+2|z|+1 \leq 3+2+1=6$

Answer 2

If $|z|\le 1$ then $$|3z^2+2z+1|\le |3z|^2+|2z|+|1|=$$ $$=|3|\cdot |z|\cdot |z|+|2|\cdot |z|+|1|\le$$ $$\le 3\cdot 1\cdot 1+2\cdot 1+1=6$$ which cannot be lowered because $|3z^2+2z+1|=6$ when $z=1.$