For $\triangle ABC$ with circumradius $R$ and nine-point center $E$, prove that $EA+EB+EC\le3R$ 
Given a triangle $ABC$ with the circumradius $R$, the centroid $G$, and the nine-point center $E$. Prove that $$EA+EB+EC\le 3R$$

Note. I'm trying to use the vector equation below  to prove this inequality, however, I'm stuck: $$\overrightarrow{EA} + \overrightarrow{EB} + \overrightarrow{EC} = 3\overrightarrow{EG}$$
 A: 
We need the help of two triangle inequalities to prove the inequality mentioned in OP’s statement. We provide the proof of one of them while leaving the other for OP to try his/her hand at.
Consider the scalene triangle $UVW$ shown in $\mathrm{Fig.\space 2}$. Its sides and one of its medians (i.e. $WM$) have lengths $u$, $v$, $w$, and $m$. We shall write,
$$u^2=m^2+\frac{w^2}{4}+m\times w\times \cos\left(\omega\right)\qquad\left(\mathrm{for}\space \triangle MVW\right), \tag{1}$$
$$v^2=m^2+\frac{w^2}{4}-m\times w\times \cos\left(\omega\right)\qquad\left(\mathrm{for}\space \triangle UMW\right). \tag{2}$$
When we add (1) to (2), we get,
$$u^2+v^2 =2m^2+\frac{w^2}{2}\quad\rightarrow\quad 4m^2= 2u^2+2v^2-w^2. \tag{3}$$
As shown below, equation (3) can be transformed into the first of the two sought triangle inequalities, which is, in fact, the mainstay of the proof of OP’s inequality.
$$4m^2= 2u^2+2v^2-w^2= u^2+v^2+\left(u^2+v^2-w^2\right) = u^2+v^2+2uv\cos\left(\phi\right)$$
Since thelargest value of $\cos\left(\phi\right)$ is 1, we cam write,
$$4m^2\le u^2+v^2+2uv=\left(u+v\right)^2 \quad\rightarrow\quad   u+v\ge 2m. \tag{4}$$
Here, the equality holds only for the degenerated triangle. The purpose of the numerical values shown in the diagram is to convince you that this inequality holds.
Now, pay your attention to the $\mathrm{Fig.\space 1}$, which depicts the configuration described by OP and auxiliary points and segments introduced by us. $H$ and $O$ are the orthocenter and circumcenter of the triangle $ABC$ respectively. We assume that OP is aware of the fact that the center of the nine-point circle $E$ is the midpoint of the segment $HO$. We denote the radius of the circumcircle $ABC$ as $R$.
Consider the triangle $AOH$, where $AE$ is one of the medians. We can apply the inequality (4) to this triangle to obtain $2AE\le AO+AH$. Since $AO=R$, this becomes $2AE\le R+AH$. In a similar vein, we can state $2BE\le R+BH$ and $2CE\le R+CH$ by considering the triangles $BHO$ and $CHO$ respectively. When we add these three inequalities together, we have,
$$2\left(AE+BE+CE\right)\le 3R+\left(AH+BH+CH\right). \tag{5}$$
Let us try to express the segment $AH_b$ as a function of $R$. Consider the isosceles triangle $ABO$. There, we have $AB=2R\sin\left(\hat{C}\right)$. Next, consider the right-angled triangle $ABH_b$. It is easy to see that $$AH_b =AB\cos\left(\hat{A}\right)= 2R\sin\left(\hat{C}\right) \cos\left(\hat{A}\right). \tag{6}$$
Finally, we consider the right-angled triangle $AHH_b$ to obtain
$$AH_b =AH\sin\left(\hat{C}\right) \tag{7}.$$
Now, we have two expressions for $AH_b$, i.e. (6) and (7). From them, it follows,
$$AH_b =2R\sin\left(\hat{C}\right) \cos\left(\hat{A}\right)= AH\sin\left(\hat{C}\right)\qquad\rightarrow\qquad AH=2R\cos\left(\hat{A}\right).$$
Similarly, we are able to express both $BH$ and $CH$ as $BH=2R\cos\left(\hat{B}\right)$ and $CH=2R\cos\left(\hat{C}\right)$ respectively. When we substitute these values in (5), we get,
$$2\left(AE+BE+CE\right)\le 3R+2R\left(\cos\left(\hat{A}\right)+\cos\left(\hat{B}\right)+\cos\left(\hat{C}\right)\right). \tag{8}$$
It is up to OP to prove using trgonometry, for a given triangle,
$$ \cos\left(\hat{A}\right)+\cos\left(\hat{B}\right)+\cos\left(\hat{C}\right)\le \frac{3}{2}.$$
Once OP has done that, he/she can insert it in (8) to get the sought inequality, i.e.
$$AE+BE+CE\le  3R, $$
where equality holds if and only if $\triangle ABC$ is an equilateral triangle, This is because, in an equilateral triangle, the two centers in question coincide.
