<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] Let $a=b=c=1$.
Thus, we obtain a value $3$.
We'll prove that it's a minimal value, for which it's enough to prove that $$a^8+b^8+c^8+2(a-1)(b-1)(c-1)\geq3.$$
We'll prove that this inequality is true even for any reals $a$, $b$ and $c$ such that $a+b+c=3$.
Indeed, by AM-GM $$\sum_{cyc}a^8\geq\sum_{cyc}(2a^4-1)$$ and it's enough to prove that:
$$2(a^4+b^4+c^4)-3+2(abc-ab-ac-bc+3-1)\geq3$$ or
$$a^4+b^4+c^4+abc-ab-ac-bc-1\geq0$$ or
$$81(a^4+b^4+c^4)+27(a+b+c)abc-9(a+b+c)^2(ab+ac+bc)-(a+b+c)^4\geq0$$ or
$$\sum_{cyc}(80a^4-13a^3b-13a^3c-24a^2b^2-30a^2bc)\geq0$$ or
$$13\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+67\sum_{cyc}(a^4-a^2b^2)+43\sum_{cyc}(a^2b^2-a^2bc)\geq0$$ or
$$13\sum_{cyc}(a-b)^2(a+b-c)^2+67\sum_{cyc}(a^2-b^2)^2+43\sum_{cyc}c^2(a-b)^2\geq0$$ and we are done!
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