if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$ This is supposed to be an application of AM-GM inequality.

if $abc=1$, then the following holds true: $a^2+b^2+c^2\ge a+b+c$

First of all,
$a^2+b^2+c^2\ge 3$
by a direct application of AM-GM.Also,we have
$a^2+b^2+c^2\ge ab+bc+ca$
Next,we consider the expression
$(a+1)(b+1)(c+1)=a+b+c+ab+bc+ca+abc\le a+b+c+a^2+b^2+c^2+1$
but that hardly helps.I know that
$3(a^2+b^2+c^2)\ge (a+b+c)^2$
From the first derived inequality,we know that the left hand side of the above inequality is greater than or equal to 9.But I can't see how that can be used here.I know that we can get $a+b+c\ge 3$ using AM-GM but that does not take me a step closer to finding the solution(from what I can understand).Also,
$a^3+b^3+c^3\ge a^2b+b^2c+c^2a$
$(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]\ge a^2b+b^2c+c^2a$
A hint will be appreciated at this point.

Since $(2,0,0)\succ\left(\frac{4}{3},\frac{1}{3},\frac{1}{3}\right)$, our inequality it's just Muirhead.
