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

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 $(a,b,c)=(1,1,1)$.
