Minimum value of $\left|z^2-z+1\right|+\left|z^2+z+1\right|$ for $z\in \mathbb{C}$ (1) If $\left|z\right| = 1$. Then find minimum value of $\left|z^2+z+4\right|$
(2) If $z\in \mathbb{C}.$ Then minimum value of $\left|z^2-z+1\right|+\left|z^2+z+1\right|.$
$\bf{My\; Try}::$ (1) Given $\left|z\right| = 1\Rightarrow z \bar{z} = 1$. 
So $\left|z^2+z+4z\bar{z}\right| = |z|\cdot \left|z+4\bar{z}+1\right| = \left|z+4\bar{z}+1\right|$
Now Let $z = x+iy$. Then $\bar{z} = x-iy$ and $|z| =1\Rightarrow x^2+y^2 = 1$
So $\left|x+iy+4x-4iy+1\right| = \left|5x+1-3iy\right| = \sqrt{(5x+1)^2+9y^2}$
So Let $f(x) = \sqrt{(5x+1)^2+9(1-x^2)} = \sqrt{25x^2+1+10x+9-9x^2}$
So $\displaystyle f(x) = \sqrt{16x^2+10x+10}=4\sqrt{x^2+\frac{5}{8}x+\frac{5}{8}}$
So $\displaystyle f(x) = 4\sqrt{\left(x+\frac{5}{16}\right)^2+\left(\frac{5}{8}-\frac{25}{256}\right)}\geq 4\sqrt{\frac{27 \times 5}{256}} = 4\times \frac{3\sqrt{15}}{16} = \frac{3\sqrt{15}}{4}$
which is occur at $\displaystyle x = -\frac{5}{16}$
(2) Now I did not understand how can i solve (II) one
Help Required
Thanks
 A: Let $z=x+yi\ (x,y\in\mathbb R)$.
Let us consider the case $|z|=r$ where $r$ is a fixed non-negative real number.
We have
$$\small\begin{align}&\left|z^2-z+1\right|+\left|z^2+z+1\right|
\\\\&=|(x+yi)^2-(x+yi)+1|+|(x+yi)^2+x+yi+1|
\\\\&=\sqrt{(x^2-y^2-x+1)^2+(2xy-y)^2}+\sqrt{(x^2-y^2+x+1)^2+(2xy+y)^2}
\\\\& \stackrel{y^2=r^2-x^2}=\sqrt{4x^2+(-2r^2-2)x+r^4-r^2+1}+\sqrt{4x^2+2(r^2+1)x+r^4-r^2+1}
\\\\&=\sqrt{\bigg(2x-\frac{r^2+1}{2}\bigg)^2+\bigg(0-\frac{(r^2-1)\sqrt 3}{2}\bigg)^2}+\sqrt{\bigg(2x-\frac{-r^2-1}{2}\bigg)^2+\bigg(0-\frac{(r^2-1)\sqrt 3}{2}\bigg)^2}\end{align}$$
This can be seen as the sum of the distance between $A$ and $B$ and the distance between $A$ and $C$ where
$$A(2x,0),\quad B\bigg(\frac{r^2+1}{2},\frac{(r^2-1)\sqrt 3}{2}\bigg),\quad C\bigg(\frac{-r^2-1}{2},\frac{(r^2-1)\sqrt 3}{2}\bigg)$$
For any fixed $r$, the minimum value of the sum is attained when $A$ is on the line passing through $B$ and $$D\bigg(\frac{-r^2-1}{2},\color{red}{-}\frac{(r^2-1)\sqrt 3}{2}\bigg)$$
The equation of the line $BD$ is given by
$$y-\frac{(r^2-1)\sqrt 3}{2}=\frac{(r^2-1)\sqrt 3}{r^2+1}\bigg(x-\frac{r^2+1}{2}\bigg)$$
So, if $A$ is on this line, then we get
$$0-\frac{(r^2-1)\sqrt 3}{2}=\frac{(r^2-1)\sqrt 3}{r^2+1}\bigg(2x-\frac{r^2+1}{2}\bigg)$$
from which $x=0$ follows.
So, all we need is to consider the case $x=0$.
For $x=0$, we get
$$|z^2-z+1|+|z^2+z+1|=\sqrt{(2y^2-1)^2+3}\ge \sqrt 3$$
Therefore, the minimum value of $|z^2-z+1|+|z^2+z+1|$ is $\color{red}{\sqrt 3}$ which is attained when $z=\pm\frac{i}{\sqrt 2}$.
A: HINT. The second part is the triangle inequality in reverse. So instead of starting with $|a+b|$ and getting $|a+b|\leq |a|+|b|$. You have $|a|+|b|$ and want to write something like $|a+b|\leq |a|+|b|$. 
A: For part 2:
Let $a = z^2 + 1$ and $b = z$. We have $a - 1 = b^2$.
We have
\begin{align*}
 &(|z^2 - z + 1| + |z^2 + z + 1|)^2\\
 =\,& |a - b|^2 + |a + b|^2 + 2|a - b|\cdot |a + b| \\
 =\,& 2(|a|^2 + |b|^2) + 2|a^2 - b^2|\\
 =\,& 2 |a|^2 + 2|a - 1| + 2|a^2 - a + 1|\\
 \ge\,& 2|a|^2 + 2\mathrm{Re}(1 - a) + 2\mathrm{Re}(a^2 - a + 1)\\
 =\,& 2|a|^2 + 2\mathrm{Re}(a^2) + 4 - 4\mathrm{Re}(a)\\
 =\,& 4[\mathrm{Re}(a)]^2 + 4 - 4\mathrm{Re}(a)\\
 =\,& [2\mathrm{Re}(a) - 1]^2 + 3\\
 \ge\,& 3
\end{align*}
where we have used $|u| \ge \mathrm{Re}(u)$,
and $|u|^2 + \mathrm{Re}(u^2) = 2[\mathrm{Re}(u)]^2$.
Also, when $z = \frac{1}{\sqrt 2}\mathrm{i}$,
we have $|z^2 - z + 1| + |z^2 + z + 1| = \sqrt 3$.
Thus, the minimum of $|z^2 - z + 1| + |z^2 + z + 1|$ is $\sqrt 3$.
