<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] Hint for $a,b,c>3$ we have :
$$\frac{a^c}{b^b}+\frac{b^c}{c^b}+\frac{c^c}{a^b}\geq \frac{(a+b+c)^{c-b}}{3^{c-b-1}}$$
And :
$$\frac{a^b}{b^a}+\frac{b^b}{c^a}+\frac{c^b}{a^a}<3\frac{c^b}{a^a}$$
We have for $c\geq 2(a+b)$:
$$\frac{(a+b+c)^{c-b}}{3^{c-b-1}}-3\frac{c^{b}}{a^{a}}>0$$
For the case $a+b\leq c\leq 2(a+b)$ :
We have :
$$\frac{c^{c}}{a^{b}}-\left(\frac{a^{b}}{b^{a}}+\frac{b^{b}}{c^{a}}+\frac{c^{b}}{a^{a}}\right)>0$$
Or :
$$\frac{c^{c}}{a^{b}}-3\frac{c^{b}}{a^{a}}>0$$
With your constraint we have :
$$f(x)=\frac{\left(x+2\right)^{\left(x+2\right)}}{x^{\left(x+1\right)}}-\frac{\left(x+2\right)^{\left(x+1\right)}}{x^{x}}\geq \lim_{x\to\infty}f(x)$$
$$g(x)=\frac{x^{\left(x+2\right)}}{\left(x+1\right)^{\left(x+1\right)}}-\frac{x^{\left(x+1\right)}}{\left(x+1\right)^{x}}\geq \lim_{x\to \infty}g(x)$$
$$h(x)=\frac{\left(x+1\right)^{\left(x+2\right)}}{\left(x+2\right)^{\left(x+1\right)}}-\frac{\left(x+1\right)^{\left(x+1\right)}}{\left(x+2\right)^{x}}\geq \lim_{x\to 0^+}h(x)$$
Reference :
https://arxiv.org/pdf/1504.05874.pdf
</s>