<s>[INST] Minimum three variable expression $\frac{(a^2+b^2+c^2)^3}{(a+b+c)^3|(a-b)(b-c)(c-a)|}$ If $a,b,c \ge 0$ are distinct real numbers, find the minimum value of
$$\frac{(a^2+b^2+c^2)^3}{(a+b+c)^3|(a-b)(b-c)(c-a)|}$$
What I did: I used $a^2+b^2+c^2\geq \frac{1}{3}(a+b+c)^2$, but no success. A friend of mine said the minimum happens when one of the variables is $0$, but I don't really understand why or how to prove this.
 [/INST] Your friend was right!
Indeed, we can assume that $a<b<c$.
Thus, it's enough to prove that
$$(a+b+c)^3(c-a)(c-b)(b-a)\leq bc(c-b)(b+c)^3$$ or
$$a((b+c)(b^3+c^3)+(b+c)(2b^2+bc+2c^2)a-bca^2-2(b+c)a^3-a^4)\geq0,$$ which is obvious.
Also, $$(a^2+b^2+c^2)^3\geq(b^2+c^2)^3$$ and after assuming $c=xb$ it's enough to find $$\min_{x>1}\frac{(x^2+1)^3}{(x+1)^3(x-1)x},$$ The rest is smooth.
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