<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] 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 $0$.
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