<s>[INST] Find the minimum value of $a^8+b^8+c^8+2(a-1)(b-1)(c-1)$ Let $a,b,c$ be the lengths of the three sides of the triangle, $a+b+c=3$. Find the minimum value of $$a^8+b^8+c^8+2(a-1)(b-1)(c-1)$$
My attempts:
$\bullet$ The minimum value is $3$, equality holds iff $a=b=c=1$ so by AM-GM, we have: $$a^8+b^8+c^8\ge 8(a+b+c)-21=3$$
$\bullet$ We need to prove $$3+2(a-1)(b-1)(c-1)\ge3$$ or $$abc-(ab+bc+ca)+a+b+c-1\ge0$$
$\bullet$ Note that $ab+bc+ca\le \dfrac{(a+b+c)^2}{3}=3 $ so we need to prove: $$abc-3+3-1\ge0$$ or $$abc\ge1$$
But I have no idea from here, please help me
 [/INST] Another way.
We'll prove that the inequality
$$a^8+b^8+c^8+2(a-1)(b-1)(c-1)\geq3$$ is true for any positives $a$, $b$ and $c$ such that $a+b+c=3$.
Indeed, since $$\prod_{cyc}((a-1)(b-1))=\prod_{cyc}(a-1)^2\geq0,$$ we can assume $a(b-1)(c-1)\geq0$ and by AM-GM and C-S we obtain:
$$\sum_{cyc}a^8+2\prod_{cyc}(a-1)-3\geq\sum_{cyc}(4a^2-3)+2\prod_{cyc}(a-1)-3=$$
$$=\sum_{cyc}(4a^2-3)+2(abc-ab-ac-bc+3-1)-3=$$
$$=3(a^2+b^2+c^2-3)+a^2+b^2+c^2+2abc+1-2(ab+ac+bc)\geq$$
$$\geq a^2+b^2+c^2+2(ab+ac-a)+1-2(ab+ac+bc)=(b-c)^2+(a-1)^2\geq0.$$
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