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
 A: The function $f(a,b,c)=a^8+b^8+c^8+2(a-1)(b-1)(c-1)$ has only one extrema minimum point $f(1,1,1)=3$ on the domain $a+b+c=3$. To prove that this one is the global minimum we need to check the value of $f$ at the boundary of the following domain:

*

*$0<a<\frac 32$

*$\frac32-a<b<\frac 32$

*$c=3-a-b$

and we have

*

*for $a\to0 \implies b,c\to\frac 32$
$$f\left(0, \frac 32, \frac 32\right)=2\left(\frac32\right)^8 -2\left(\frac12\right)^2>3$$

*

*for $a\to\frac32 \implies 0<b<\frac 32$ and $c=\frac 32-b$ we obtain
$$f\left(\frac32, b, \frac32-b\right)=\left(\frac32\right)^8+b^8+\left(\frac32-b\right)^8+2\left(\frac12\right)\left(b-1\right)\left(\frac12-b\right)>3$$
which by symmetry suffices, therefore the minimum value is attained at $(a,b,c)=(1,1,1)$.
A: 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:
$$\dfrac{a^8+b^8+c^8}{3}\geq\left(\dfrac{a^4+b^4+c^4}{3}\right)^2\geq\dfrac{a^4+b^4+c^4}{3}\cdot\left(\dfrac{a+b+c}{3}\right)^4=\dfrac{a^4+b^4+c^4}{3}$$
A: 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!
A: 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:
$$\sum_{cyc}a^8+2\prod_{cyc}(a-1)-3\geq\sum_{cyc}(4a^2-3)+2\prod_{cyc}(a-1)-3=$$
$$=\sum_{cyc}(4a^2-3)+2(abc-ab-ac-bc+3-1)-3=$$
$$=3(a^2+b^2+c^2-3)+a^2+b^2+c^2+2abc+1-2(ab+ac+bc)\geq$$
$$\geq a^2+b^2+c^2+2(ab+ac-a)+1-2(ab+ac+bc)=(b-c)^2+(a-1)^2\geq0.$$
