<s>[INST] Prove that $a^2+b^2+c^2\geq [2(a-b)^2(b-c)^2(a-c)^2]^{1/3}$ Mathematica seems to know that this statement is true, yet I am struggling to prove it. Possible useful inequalities are Minkowski and the geometric mean. Using the geometric mean inequality I can prove a less strict bound:
$$(a-b)^2+(b-c)^2+(a-c)^2 \geq 0 \implies 4 a^2+4 b^2+4 c^2 \geq (a-b)^2+(b-c)^2+(a-c)^2,$$ 
then the geometric mean inequality gives
$$ (a-b)^2+(b-c)^2+(a-c)^2 \geq 3[(a-b)^2(b-c)^2(a-c)^2]^{1/3} .$$ 
Combining the two we get
$$  a^2+b^2+c^2 \geq  \frac{3}{4}[(a-b)^2(b-c)^2(a-c)^2]^{1/3},$$ 
which is not strong enough.
Can anyone prove this for $2^{1/3}$ in place of $3/4$?
 [/INST] Here is another approach that might shed a bit more light on the essence of the inequality.

The AM-GM inequality gives
$$
\begin{align}
a^2+b^2+c^2
&=\frac{(a-b)^2+(b-c)^2+(c-a)^2+(a+b+c)^2}3\\
&\ge\frac{(a-b)^2+(b-c)^2+(c-a)^2}3\\
&\ge\left[(a-b)^2(b-c)^2(c-a)^2\right]^{1/3}\tag{1}
\end{align}
$$
Equality is attained in the second inequality of $(2)$ only when $(a-b)^2=(b-c)^2=(c-a)^2$, which cannot happen since $(a-b)+(b-c)+(c-a)=0$. This indicates that a better inequality may be possible.

If we set $x=a-b$ and $y=b-c$, we can look for the maximum of
$$
\frac{x^2y^2(x+y)^2}{\left(x^2+y^2+(x+y)^2\right)^3}=\frac{z^2(z+1)^2}{\left(z^2+1+(z+1)^2\right)^3}\tag{2}
$$
where $z=x/y$ ($y=0$ is certainly not the maximum). The derivative of the log of $(2)$ is
$$
\frac{2(2z+1)}{z(z+1)}-\frac{6(2z+1)}{z^2+1+(z+1)^2}=\frac{(2+z)(1-z)(1+2z)}{z(1+z)(1+z+z^2)}\tag{3}
$$
which vanishes where $z=-\frac12$, $z=1$, and $z=-2$. Each gives a value of $\frac1{54}$ in $(2)$. Therefore,
$$
\left(x^2+y^2+(x+y)^2\right)^3\ge54x^2y^2(x+y)^2\tag{4}
$$
Substituting back for $x$ and $y$ in $(4)$ gives
$$
\frac{(a-b)^2+(b-c)^2+(c-a)^2}3\ge\left[2(a-b)^2(b-c)^2(c-a)^2\right]^{1/3}\tag{5}
$$

To get the best inequality, we can use $(5)$ instead of the AM-GM:
$$
\begin{align}
a^2+b^2+c^2
&=\frac{(a-b)^2+(b-c)^2+(c-a)^2+(a+b+c)^2}3\\
&\ge\frac{(a-b)^2+(b-c)^2+(c-a)^2}3\\[6pt]
&\ge\left[2(a-b)^2(b-c)^2(c-a)^2\right]^{1/3}\tag{6}
\end{align}
$$
</s>