How prove this $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\ge |(a-1)(b-1)(c-1)|+a+b+c$ let $a,b,c$ are positive numbers,and  such $abc\le 1$,prove that
$$\dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b}\ge |(a-1)(b-1)(c-1)|+a+b+c$$
 A: case 1:
For case $a<1,b<1,c<1$, it is trivial the inequality work(Sudeep's comments)
case 2:
now we prove when one of $a,b,c$ equals $1$,the inequality also works. 
WLOG,Let $c=1$, the inequality $\iff b^2+a \ge ab^2+ab$, let $ab=k \le 1 \iff b^3-kb^2-kb+k \ge0 \iff b^3(1-k)+k(b-1)^2(b+1)\ge0$ 
so the "=" will hold when $k=1,b=1 \implies a=b=c=1$
we will prove the case $abc=1$ first:
there are two cases:
case 3: two of them $\ge 1$  , the inequality $\iff  \dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b}\ge ab+bc+ac \iff a^3b^3-a^2b^3+b^3-b^2-ab+1\ge 0 \iff $
$b^3(a-1)^2(a+1)+(b^2-1)(a-1) \ge 0$ when $a,b$ both $\ge 1 $
$b(ab-1)^2(ab+1)+b^2(1-b)(a^2-1) \ge 0 $ when one of $a,b \ge1$ and another $\le1$
case 4: one of them $\ge1$, the other two $\le1$ ,the inequality $\iff  \dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b}\ge 2(a+b+c)-(ab+bc+ac) \iff a^3b^3+a^2b^3+b^3+b^2+1-2ab^3-2a^2b^2-2b \ge0 \iff (b-1)^2+ab(ab-1)^2+b^3(a-1)^2 \ge 0$
now we discuss $a'b'c'<1$ cases:
let $a'b'c'=k^3,a'=ka,b'=kb,c'=kc,abc=1$, we only consider RHS becasue LHS will be same.
denote $H$ for $a,b,c ,H'$ for $a',b',c'$,
if $H'$ in case 3,then $H$ will be in case 3 also:
$H-H'=(ab+bc+ac)(1-k^2)>0$
if both $H'$ and $H $ in case 4:
$H-H'=2(1-k)(a+b+c)-(1-k^2)(ab+bc+ac)+1-k^3$, let $c \ge 1 \iff $
$2(1-k)(a^2b+b^2a+1)-(1-k^2)(a^2b^2+a+b)+(1-k^3)ab =(1-k)((2ab-(1+k))(a+b)+2)+(1-k)(1+k)(ab-a^2b^2)+(1-k)k^2ab \ge 0 \implies H \ge H'$
if $H'$ in case 4, but $H$ in case 3, then there is always one case that $a=ka'>1$ suppose $c \ge 1, a\ge b$ we can chose $k_2=\dfrac{1}{a'}$,then $a=1,b\le1, c\ge 1$ which make $H$ still in case 4. and $H$ now is belong to case 2.So the inequality is also true.
QED. 
