Problem from Algebraic representation of complex numbers

Question

Prove that

$$√(7/2)≤|1+z|+|1-z+z²|≤3√(7/6)$$

For all complex numbers with $|z|=1$.

I proved $|1+z|+|1-z+z²|≤3$ so the right hand inequality

But didn't proved left hand inequality

Answer 1

Hint:  with $z = e^{i\,2 \varphi}\,$:

• $\displaystyle |1+z| = 2 |\cos \varphi|\,$;

• using the above, $\require{cancel}\displaystyle \left|1-z+z^2\right|=\left|\frac{1+z^3}{1+z}\right|=\frac{\cancel{2} \left|\cos 3\varphi\right|}{\cancel{2} |\cos \varphi|}=\left|4 \cos^2 \varphi-3\right|\,$.

Let $\,t = |\cos \varphi|\,$, then the problem reduces to finding the range of $\,2 t + \left|4t^2-3\right|\,$ for $\,t \in [0,1]\,$.