Minimum of $E(z)=|z+1|+|z^{2n}+1|$ 
Let $n \in \mathbb{N}$. Find the minimum of the expression $E(z)=|z+1|+|z^{2n}+1|$ over $\mathbb{C}$.

I found this problem in a Romanian magazine with all sorts of math problems.
(Edit: ${\color{blue}{\textrm{This was suggested for a contest for 10th graders}}}$, According to @alexanderv's comment.)
What I've done so far(WRONG):
$E \geq 0$ so if we can find $z$ with $E(z)=0$ it's enough. If $n=0$, the minimum is achieved in $-1$, $E(z) =2$. If $n \geq 1$, let's take $z= \cos \theta + i\sin \theta$. Using $|1+\cos \alpha + i\sin \alpha|=2| \cos (\frac{\alpha}{2})|$, we get $E(z)=2(|cos(\frac{\theta}{2})|+|cos(n\theta)|)$. Then, taking the modulus and using $\cos(\pi-\alpha)=-\cos(\alpha)$ and the $\cos(x) + \cos(y)$ formula, we take $z$ so one of the product terms equals to $0$, achieving the minimum. But I'm not sure if it's right, can someone approve?
What's wrong above is $E(z)=0$ implies $z=-1$, that means $E(z)=2$, contradiction. I don't have any idea how to find the minimum.
 A: $\def\z{\bar z}$
Let $$E_m(z)=|z+1|+|z^{m}+1|;\quad m\in2\mathbb Z_+.\tag1
$$
It is convenient to rewrite the equation $(1)$ as:
$$
E_m(z,\z)=\sqrt{(z+1)(\z+1)}+\sqrt{(z^m+1)(\z^m+1)},\tag2
$$
where $\z$ is the complex conjugate of $z$, and treat $z$ and $\z$ as independent variables (they are of course not independent but this is a well-known and very convenient tool for finding the critical points). Obviously the function $E_m(z,\z)$ is differentiable at all points $(z,\z)$ unless one of the square roots is $0$.
As easy to check among $m$ solutions of the equation $z^{m}+1=0$ the root $z=e^{i\frac{m-1}{m}\pi}$ delivers the least value of $E_m$ which reduces to
$$|z+1|=2\sin\tfrac\pi{2m}.\tag3$$
We claim that $(3)$ is the global minimum of $E_m(z)$. To prove this it suffices to show that the value of the function $E_m(z)$ at all stationary points is higher than that given by $(3)$.
The stationary points can be determined from the equations:
$$\begin{align}
&\frac{\partial E_m}{\partial z}=0
\implies
\frac{(\z^m+1)m z^{m-1}}{\sqrt{(z^m+1)(\z^m+1)}}=
-\frac{\z+1}{\sqrt{(z+1)(\z+1)}};\tag{4a}\\
&\frac{\partial E_m}{\partial \z}=0
\implies
\frac{(z^m+1)m \z^{m-1}}{\sqrt{(z^m+1)(\z^m+1)}}=
-\frac{z+1}{\sqrt{(z+1)(\z+1)}}.\tag{4b}\\
\end{align}$$
Multiplying the corresponding sides of the equations one obtains:
$$
m^2|z|^{2(m-1)}=1\implies |z|=m^{-\frac1{m-1}}\equiv\rho_m.\tag5
$$
The equation $(5)$ means that the stationary points $z_s$ and their images $z_s^m$ lie on the concentric circles with center at $(0,0)$ and radii $\rho_m$ and $\rho_m^{m}$, respectively ($\rho_m^{m}<\rho_m<1$). This finding helps to visualize the problem since the function $E_m(z_s)$ is geometrically the sum of distances from the point $z=-1$ to $z_s$ and $z_s^m$, which is bounded from below by the value:
$$
(1-\rho_m)+(1-\rho^{m}_m)=2-\left(1+\frac1{m}\right)\rho_m>1.\tag6
$$
The last inequality for $m>1$ follows most simply from $x\log(x)+(1-x)\log(1+x)<0$ valid for all $x:\ 0<x<1$.
Since $2\sin\frac\pi{4n}<1$ for $n\ge2$ it remains to consider only the case $n=1$. For this we explicitly find the solutions of $(4)$ for $m=2$. To facilitate the computation we substitute $z=\rho_2e^{i\phi}=\tfrac12e^{i\phi}$ in one of the equations. After straightforward algebra one finds the solutions $z_1=-\tfrac12$, $z_{2,3}=\tfrac{-3\pm i\sqrt7}8$ and determines:
$$E_2(z_{2,3})=\tfrac54\sqrt2>  E_2(z_1)=\tfrac74>2\sin\tfrac\pi4=\sqrt2,$$
which finalizes the proof.
A: [Update 09 April 2021] Here is a partial proof for the case $|z| \ge 1$.
We have with $z = r e^{-i \phi}$:
$$\begin{align}
E(z)=&|z+1|+|z^{2n}+1| = |r e^{-i \phi}+1|+|r^{2n} e^{-i 2 n \phi}+1| \\
=& |r + e^{i \phi}|+|r^{2n} + e^{i 2 n \phi}| \\
=& \sqrt{(r + \cos\phi)^2 + \sin^2\phi} + \sqrt{(r^{2n} + \cos(2n\phi))^2 + \sin^2(2n\phi)}\\
=& \sqrt{r^2 - 2 r  +1 + 2 r (1+\cos\phi)} + \sqrt{r^{4n} - 2 r^{2n}  +1 + 2 r^{2n} (1+\cos(2n \phi))} \\
=& \sqrt{(r-1)^2 + 4 r \cos^2(\phi/2)} + \sqrt{(r^{2n} -1)^2 + 4 r^{2n}  \cos^2( n \phi)} \\
\ge& 2 \sqrt{ r} |\cos(\phi/2)| + 2 r^n |\cos(n \phi)| \qquad {\rm\bf [1]}\\
\ge& 2  |\cos(\phi/2)| + 2  |\cos( n \phi)| \qquad {\rm\bf [2]}
\end{align}$$
Note in ${\rm\bf [1]}$ that this step is tight for $r=1$ and in ${\rm\bf [2]}$ that both $\sqrt{r}$ and $r^n$ are  rising functions, so, whatever $\phi$,  the smallest value is given for $r=1$ (in the considered range $r \ge 1$), so this is again tight. So the last result is indeed the correct minimum for $r \ge 1$.
The last result has now to be minimized w.r.t. $\phi$.  W.l.og. we can take $0\le \phi\le \pi$ so we have to minimize $2 \cos(\phi/2) + 2  |\cos( n \phi)|$. Since $\cos(\phi/2)$ is falling for $0\le \phi\le \pi$, the minimum will be attained closest to $\pi$ at a point where  $\cos( n \phi)$ changes sign, i.e. at $\phi = (\frac12 + k) \pi/n $ with the integer $k$ chosen such that $\phi$ comes closest to $\pi$, i.e. $k = n-1$. This gives $\phi = \pi \frac{2n-1}{2n}$ and hence the minimum for $|z| \ge 1$ is obtained as
$$E(z) \ge  2  \cos(\pi \frac{2n-1}{4n}) = 2  \sin(\frac{\pi}{4n}) $$
What remains to be shown is that for $|z| <1$, no smaller values of $E(z)$ are attained. The above method does not work, as the step  ${\rm\bf [1]}$ above neglects the terms $(r-1)^2$ and $(r^{2n}-1)^2$ which are falling for $r<1$. Hence as $r$ moves away from $1$ (i.e. gets smaller), the true value of $E(z)$ is increased by these terms, so the bound ${\rm\bf [1]}$ used above gives way too small values for $r<1$. However, numerical evidence indeed supports that the minimum  of $E(z)$ is attained at $|z| =1$, so in this case the derived minimum is indeed valid for all $z$.
I didn't succeed to do that part of the proof yet, suggestions are welcome.
A: shymilan@AoPS's elegant proof (I rewrote it):
Let $w_k = \mathrm{e}^{\mathrm{i}\frac{(2k - 1)}{2n}\pi}, ~ k = 1, 2, \cdots, 2n$. Then $Q(z) := z^{2n} + 1
= (z - w_1)(z - w_2)\cdots (z - w_{2n})$.
We have the partial fraction decomposition
$$\frac{1}{z^{2n} + 1} = \frac{1}{Q(z)}
= \sum_{k=1}^{2n} \frac{1}{Q'(w_k)}\frac{1}{z - w_k}
= \sum_{k=1}^{2n} \frac{-w_k}{2n}\frac{1}{z - w_k}. \tag{1}$$
(easy to prove, e.g. see Partial_fraction_decomposition).
For $k = 1, 2, \cdots, 2n$, we have
\begin{align*}
 |z + 1| + |z^{2n} + 1| &= 
 |z - w_k + w_k + 1| + |z-w_k| \cdot \left|\frac{z^{2n} + 1}{z - w_k}\right|\\
 &\ge |w_k + 1| - |z - w_k|
 +  |z-w_k| \cdot \left|\frac{z^{2n} + 1}{z - w_k}\right|\\
 &= |w_k + 1| + |z - w_k|
 \left(\left|\frac{z^{2n} + 1}{z - w_k}\right| - 1\right). \tag{2}
\end{align*}
Using (1), we have
\begin{align*}
 \sum_{k=1}^{2n} \left(\left|\frac{z^{2n} + 1}{z - w_k}\right| - 1\right)
 &= \sum_{k=1}^{2n} \left|\frac{z^{2n} + 1}{z - w_k}\right| - 2n \\
 &= \sum_{k=1}^{2n} \left|\frac{z^{2n} + 1}{z - w_k} \cdot w_k\right| - 2n\\
 &\ge \left|\sum_{k=1}^{2n} \frac{z^{2n} + 1}{z - w_k} \cdot w_k\right| - 2n\\
 &= 0. \tag{3}
\end{align*}
From (2) and (3), there exists $k \in \{1, 2, \cdots, 2n\}$ such that $\left|\frac{z^{2n} + 1}{z - w_k}\right| - 1 \ge 0$ and
$$|z + 1| + |z^{2n} + 1| \ge |w_k + 1|.$$
For $k = 1, 2, \cdots, 2n$, we have
$$|w_k + 1| = |\mathrm{e}^{\mathrm{i}\frac{(2k - 1)}{2n}\pi} + 1| = 2\left|\sin \frac{(2n - 2k + 1)\pi}{4n}\right|
\ge 2 \sin \frac{\pi}{4n}.$$
Thus, we have
$$|z + 1| + |z^{2n} + 1| \ge 2 \sin \frac{\pi}{4n}.$$
Also, if $z = \mathrm{e}^{\mathrm{i}\frac{2n - 1}{2n}\pi}$,
we have $|z+1| + |z^{2n} + 1| = 2 \sin \frac{\pi}{4n}$.
Thus, the minimum of  $|z+1| + |z^{2n} + 1|$ is $2 \sin \frac{\pi}{4n}$.
