For $c>b>a$ with $c,b,a\in\Bbb N,c-b=b-a=1$, prove $\frac{a^c}{b^b}+\frac{b^c}{c^b}+\frac{c^c}{a^b}\ge\frac{a^b}{b^a}+\frac{b^b}{c^a}+\frac{c^b}{a^a}$

Question

Given $c>b>a$, also $c-b=b-a=1$ where $c,b,a$ are Natural numbers, prove that $$ \frac{a^c}{b^b}+\frac{b^c}{c^b}+\frac{c^c}{a^b} \geqslant \frac{a^b}{b^a}+\frac{b^b}{c^a}+\frac{c^b}{a^a} $$

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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.

Answer 1

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.

Answer 2

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