<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] WLOG assume $a\leq 1.$ Then, let $f(a,b,c) = a^8+b^8+c^8 +2(a-1)(b-1)(c-1)$ and $\dfrac{b+c}{2} = t$ consider:
$$f(a,b,c) - f\left(a,t,t\right) = \left(b^8+c^8 - 2\left(\dfrac{b+c}{2}\right)^8\right)+\dfrac{1}{2}(1-a)(b-c)^2\geq 0.$$
This reduces your inequality into single variable case:
$$g(t) = (3-2t)^8+2t^8-4(t-1)^3-3\geq 0$$
which should be bashable.
EDIT: In fact, it turns out that:
$$g(t)\geq (3-2t)^4+2t^4 - 4(t-1)^3-3 = 2(t-1)^2(9t^2-32t+41)\geq 0,$$
so your inequality even holds with $a^4,b^4,c^4$ etc instead of the exponent $8.$ To see why it is stronger, one can simply observe that:
$$\dfrac{a^8+b^8+c^8}{3}\geq\left(\dfrac{a^4+b^4+c^4}{3}\right)^2\geq\dfrac{a^4+b^4+c^4}{3}\cdot\left(\dfrac{a+b+c}{3}\right)^4=\dfrac{a^4+b^4+c^4}{3}$$
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