Inequality in triangle involving medians Let $ABC$ be a triangle and $M$ a point on $(BC)$, $N$ a point on $(CA)$ and $P$ a point on $(AB)$ such that the triangles $ABC$ and $MNP$ have the same centroid.
Does the following inequality hold: $$AM^2+BN^2+CP^2\ge AA'^2+BB'^2+CC'^2 ?$$ where $A',B'$ and $C'$ are the midpoints of $(BC), (AC)$ and $(AB)$.
EDIT: $AA'^2+BB'^2+CC'^2=3/4(AB^2+BC^2+CA^2)$ and from MB/MC=NC/NA=PA/PB follows that the triangles ABC and MNP have the same centroid, by Pappus Theorem.

If $MNP$ has the same centroid than $ABC$ then $$\frac{M+N+P}3=\frac{A+B+C}3$$ (if seen each point as a vector in $\mathbb R$.)
Also, as $M$ is in $BC$ then $$M=mB+(1-m)C$$ for some $0\le m\le1$.  Analogous:
$$N=nC+(1-n)A,$$
$$P=pA+(1-p)B.$$
This means that:
$$N+M+P=(1+p-n)A+(1+m-p)B+(1+n-m)C$$
Where $A+B+C=N+M+P$, therefor
$$0=(p-n)A+(m-p)B+(n-m)C$$
Having $\lambda=p-n$ and $\mu=m-p$, then $n-m=-(\lambda+\mu)$, and we get
$$0=\lambda(A-C)+\mu(B-C).$$
Unless $ABC$ are colinear, this only happens when $\lambda=\mu=0$ which means $m=n=p$.
So:
\begin{align}
M &= mB+(1-m)C \\
M-A &= mB+C-mC-A \\
M-A &= m(B-C)+(C-A) \\
AM^2=(M-A)(M-A) &= m^2(B-C)(B-C)+m(B-C)(C-A)\\&\qquad+m(C-A)(B-C)+(C-A)(C-A)^2 \\
&= m^2CB^2-2mCB\,AC\cos C+AC^2
\end{align}
By law of cosins: $AB^2=AC^2+CB^2-2CB\,AC\cos C$, therefor $2CB\,AC\cos C = AC^2+CB^2-AB^2$, replacing we have:
\begin{align}
AM^2&=m^2CB^2-m(AC^2+CB^2-AB^2)+AC^2
\\&= mAB^2 + (m^2-m)CB^2 + (1-m)AC^2
\\\text{analogously}\\
BN^2&= mBC^2 + (m^2-m)AC^2 + (1-m)BA^2\\
CP^2&= mCA^2 + (m^2-m)BA^2 + (1-m)CB^2\\
AM^2+BN^2+CP^2 &= (1-m+m^2)(AB^2 + BC^2 + CA^2)
\end{align}
Where the minimum value of $1-m+m^2$ is where $m=\frac12$ as can be prove with a little calculus.  And $m=\frac12$ is the value for $M=A',N=B',P=C'$.
