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 $3/4$?
 A: Hint: 
WOLOG, suppose $a>b>c$. Let $b=c+y$ and $a=c+y+x$. One needs to show
$$
(c+x+y)^2+(c+y)^2+c^2 \ge (2x^2y^2(x+y)^2)^{\frac{1}{3}}.
$$
Notice that $(c+x+y)^2+(c+y)^2+c^2$ attains minimal when $c=-\frac{x+2y}{3}$. It remains to show 
$$
(x+2y)^2+(x-y)^2+(2x+y)^2 \ge 9 (2x^2y^2(x+y)^2)^{\frac{1}{3}},
$$
or 
$$
x^2+y^2+xy\ge 3 (xy\frac{x+y}{2})^{\frac{2}{3}}.
$$
for positive $x, y$. Finally, by rearranging we reach
$$
x^2 + \frac{y(x+y)}{2}+ \frac{y(x+y)}{2} \ge 3 (xy\frac{x+y}{2})^{\frac{2}{3}},
$$
which is true due to the geometric mean inequality.
A: Let $x=a-b$ and $y=b-c$. Then $a-c=x+y$. Thus, we are left with
$$
(x+y+c)^2+(y+c)^2+c^2\ge(2x^2y^2(x+y)^2)^{1/3}\tag{1}
$$
To find the minimum of the left-hand side of $(1)$, we set the derivative with respect to $c$ to $0$:
$$
2(x+y+c)+2(y+c)+2c=0\tag{2}
$$
$(2)$ says that $c=-\frac{x+2y}3$. Plugging this into the left-hand side of $(1)$ yields
$$
\begin{align}
(x+y+c)^2+(y+c)^2+c^2
&=\left(\frac{2x+y}3\right)^2+\left(\frac{\vphantom{2}y-x}3\right)^2+\left(\frac{x+2y}3\right)^2\\[6pt]
&=\frac23\left(x^2+xy+y^2\right)\tag{3}
\end{align}
$$
Thus, inequality $(1)$ is equivalent to the two variable inequality
$$
\frac4{27}\left(x^2+xy+y^2\right)^3\ge x^2y^2(x+y)^2\tag{4}
$$
$(4)$ is obvious if $y=0$, so let $z=\frac xy$ and divide $(4)$ by $y^6$. We then get that $(1)$ is equivalent to the one variable inequality
$$
\frac4{27}\left(z^2+z+1\right)^3\ge z^2(z+1)^2\tag{5}
$$
and
$$
\frac4{27}\left(z^2+z+1\right)^3-z^2(z+1)^2=\frac1{27}(2z+1)^2(z-1)^2(z+2)^2\ge0\tag{6}
$$
Thus, $(1)$ is true; and using $(2)$ and $(6)$, we can determine that equality holds precisely when
$$
\overbrace{a+b+c=0}^{\large c=-\frac{x+2y}3}\qquad\text{and}\qquad\frac{a-b}{b-c}\in\left\{-\frac12,1,-2\right\}\tag{7}
$$
which is when one of $a$, $b$, or $c$ is $0$ and the sum of the other two is $0$.
A: 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:
$$
\begin{align}
a^2+b^2+c^2
&=\frac{(a-b)^2+(b-c)^2+(c-a)^2+(a+b+c)^2}3\\
&\ge\frac{(a-b)^2+(b-c)^2+(c-a)^2}3\\[6pt]
&\ge\left[2(a-b)^2(b-c)^2(c-a)^2\right]^{1/3}\tag{6}
\end{align}
$$
