Prove this inequality $a^{\frac{a}{b}}b^{\frac{b}{c}}c\geq1$ Please help me to prove this inequality.
Assume $a,b,c>0$ and $abc\geq1$ then $a^{\frac{a}{b}}b^{\frac{b}{c}}c\geq1$.
Thanks.
 A: It seems that the inequality doesn't hold. For instance, take $a = e^{-1}$, $b=10^{-5}$, and $c=\frac{1}{ab}$. Then $a^{a/b}b^{b/c}c$ would be close to zero. 
Think of the case $a=e^{-1}$, $c=\frac{e}{b}$ and $b\to 0^+$. We would like to have $$S = \frac{a}{b}\log a +\frac{b}{c}\log b + \log c \geq 0.$$ However, $S = -\frac{1}{be}+\frac{b^2}{e}\log b-1-\log b$ and as $b\to 0^+$ the first term dominates and $S\to -\infty$.
A: I think the right  inequality is:
$a^{\frac{a}{b}}b^{\frac{b}{c}}c^{\frac{c}{a}} \ge 1$
to prove it is to prove $\dfrac{a}{b}\log{a}+\dfrac{b}{c}\log{b}+\dfrac{c}{a}\log{c} \ge 0$
let $abc=1 \to a=\dfrac{1}{bc}$
LHS=$(b^3-1)\log{b}+(b^3c^3-1)\log{c}$
it is trivial when $b>1$ and $c>1$,  $b<1$ and $c<1$, LHS $> 0$
when $b<1, c>1$, if $b^3c^3 >1$, LHS $>0$
so we only check $b^3c^3<1$, ie ,$1<c<\dfrac{1}{b}$ , in this case, 
$0<\log{c} <-\log{b},0<1-b^3c^3<1-b^3 \implies (b^3c^3-1)\log{c}>(1-b^3)\log{b} \implies $ LHS$>0$
with same mehtod, we can also get when $b>1, 1>c>\dfrac{1}{b}$, LHS$>0$, when $b>1, c<\dfrac{1}{b}$, it is trivial LHS$>0$  
since $(b^3-1)\log{b}\ge 0$, when and only when $b=1$ the "$=0$" holds, if LHS $=0 \implies b=1, c=1 \implies a=1$
it is not difficult to prove when $abc>1,a^{\frac{a}{b}}b^{\frac{b}{c}}c^{\frac{c}{a}}>x^{\frac{x}{y}}y^{\frac{y}{z}}z^{\frac{z}{x}}$,where $xyz=1 $  and $a=px,b=py,c=pz, abc=p^3xyz,p>1$, 
