<s>[INST] For $c>b>a$ with $c,b,a\in\Bbb N,c-b=b-a=1$, prove $\frac{a^c}{b^b}+\frac{b^c}{c^b}+\frac{c^c}{a^b}\ge\frac{a^b}{b^a}+\frac{b^b}{c^a}+\frac{c^b}{a^a}$ 
Given $c>b>a$, also $c-b=b-a=1$ where $c,b,a$ are Natural numbers, prove that
$$
\frac{a^c}{b^b}+\frac{b^c}{c^b}+\frac{c^c}{a^b} \geqslant \frac{a^b}{b^a}+\frac{b^b}{c^a}+\frac{c^b}{a^a}
$$


The LHS becomes $$
\frac{a^{b+c} c^b+b^{b+c} a^b+c^{b+c} \cdot b^b}{(a b c)^b},
$$
and the RHS becomes
$$
\frac{a^{b+a} c^a+b^{a+b} a^a+c^{a+b} b^a}{(a b c)^a}.
$$
Now, $a^{b+c}>a^{b+a}$,$c^b>c^a$, $b^{b+c}>b^{b+a}$,$a^b>a^a$,
$c^{b+c}>c^{b+a}$,$b^b>b^a$. So the numerator on LHS is clearly bigger than the numerator on RHS. But $(abc)^{b}>(abc)^{a}$. How do I proceed.
$\textbf{Edit}$: added an important detail.
 [/INST] For $(a, b, c) = (x, x+1, x+2)$ and positive real $x$ is your inequality equivalent to
$$
 2 \frac{(x+2)^{x+1}}{x^{x+1}} - \frac{x^{x+1}}{(x+1)^{x+1}} - \frac{(x+1)^{x+1}}{(x+2)^{x+1}} \ge 0
$$
or
$$
 2 \ge \left( \frac{x^2}{(x+1)(x+2)}\right)^{x+1} + \left( \frac{x(x+1)}{(x+2)^2}\right)^{x+1}
$$
and that is true because both terms on the right are less than one.
</s>