Geometric inequality regarding a tetrahedron The circumradius of a tetrahedron $ ABCD$ is $ R$, and the lenghts of the segments connecting the vertices $ A,B,C,D$ with the centroids of the opposite faces are equal to $ m_a,m_b,m_c$ and $ m_d$, respectively. Prove that:
$ m_a+m_b+m_c+m_d\leq \dfrac{16}{3}R$
 A: Let $\vec{x}_a, \vec{x}_b, \vec{x}_c, \vec{x}_d$ be the vertices of $ABCD$.
WOLOG, we assume $|\vec{x}_a| = |\vec{x}_b| = |\vec{x}_c| = |\vec{x}_d| = R$. i.e. the circumsphere of $ABCD$ is the sphere centered at origin with radius $R$.
Let $\vec{p} = \frac14 ( \vec{x}_a + \vec{x}_b + \vec{x}_c + \vec{x}_d )$ be the center of mass of $ABCD$. Length like $m_a$ is the distance between $\vec{x}_a$ and $\frac13 ( \vec{x}_b + \vec{x}_c + \vec{x}_d )$ and hence
$$m_a = |\vec{x}_a - \frac13 ( \vec{x}_b + \vec{x}_c + \vec{x}_d )| = \frac43 | \vec{x}_a - \vec{p}|$$
Similarly, we have $m_b = \frac43 |\vec{x}_b - \vec{p}|$, $m_c = \frac43 |\vec{x}_c - \vec{p}|$ and $m_d = \frac43 |\vec{x}_d - \vec{p}|$. As a result,
$$m_a + m_b + m_c + m_d 
= \frac43 \left(|\vec{x}_a - \vec{p}| + |\vec{x}_b - \vec{p}| + |\vec{x}_c - \vec{p}| + |\vec{x}_a - \vec{p}| \right)$$
By Cauchy-Schwarz inequality,
$$\left(|\vec{x}_a - \vec{p}| + |\vec{x}_b - \vec{p}| + |\vec{x}_c - \vec{p}| + |\vec{x}_a - \vec{p}| \right)^2 \le 4 
\left(|\vec{x}_a - \vec{p}|^2 + |\vec{x}_b - \vec{p}|^2 + |\vec{x}_c - \vec{p}|^2 + |\vec{x}_a - \vec{p}|^2\right)$$
By direction computation,
$$\begin{align}
&|\vec{x}_a - \vec{p}|^2 + |\vec{x}_b - \vec{p}|^2 + |\vec{x}_c - \vec{p}|^2 + |\vec{x}_a - \vec{p}|^2\\
= & |\vec{x}_a|^2 + |\vec{x}_b|^2 + |\vec{x}_c|^2 + |\vec{x}_d|^2 - 4 |\vec{p}|^2\\
\le & 4 R^2
\end{align}$$
As a result, we get
$$m_a + m_b + m_c + m_d \le \frac43 \sqrt{4 \times 4R^2} = \frac{16}{3} R$$.
