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I would like to get some help with the following inequality where $a$ , $b$ and $c$ are positive integers.

$$a^{2a}b^{2b}c^{2c}\ge a^{b+c}b^{c+a}c^{a+b}$$

I don't know whether it's symmetric while I know it is cyclic, do we have the right to assume $a\ge b \ge c $ if it's just cyclic ?

I would also know the difference between those terms "cyclic" and "symmetric" relative to inequalities , do one term imply the other or none implies the other.

If we have the right to assume the presaid order then the inequality re-writes as

$$\left(\dfrac{a}{b}\right)^{a-b}\left(\dfrac{a}{c}\right)^{a-c}\left(\dfrac{b}{c}\right)^{b-c} \ge 1$$

which is true.

Thanks for any clarifications

Karim Kamil
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    also https://math.stackexchange.com/questions/1073068/prove-that-aabbcc-ge-abcabc-3?noredirect=1&lq=1 – Albus Dumbledore Nov 23 '21 at 12:23
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    Unsure if following approach, which seems to work, already covered in comments. Set a = rb, c = sb, eliminating all variables but $b,r,s$. I then experimented with $r = sk~: k \geq 1 \implies r \geq s$, and found that the inequality held. I am guessing that the mirror assumption that $s \geq r$ also works. – user2661923 Nov 23 '21 at 12:39
  • @MartinR I think so, thanks – Karim Kamil Nov 23 '21 at 12:41

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