<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] 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.
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