Prove that $a^ab^bc^c\ge (abc)^{(a+b+c)/{3}}$

Question

Prove that $$a^ab^bc^c\ge (abc)^{(a+b+c)/3}$$

My attempt: $$a^ab^bc^c\ge (abc)^{(a+b+c)/{3}}\implies \bigg(\dfrac{a}{b}\bigg)^{(a-b)/{3}}\bigg(\dfrac{b}{c}\bigg)^{(b-c)/{3}}\bigg(\dfrac{c}{a}\bigg)^{(c-a)/{3}}\ge 1$$

I have a slightest hint that symmetry might be helpful here, but wondering how. Please help.

I know that this question has been asked before, but I could not see any soltuion having my approach, so I am wondering if any solution is possible using my approach or it is a dead end.

http://math.stackexchange.com/questions/109783/prove-aabbcc-ge-abc-fracabc3-for-positive-numbers — miracle173, Dec 18 '14 at 07:51
@miracle173 isn't helpful, please read my question properly. Thank you. — Swadhin, Dec 18 '14 at 07:52
That was not intended as a help for you. you mentioned that the question was already posed but did not supply a link. — miracle173, Dec 18 '14 at 07:56

score 8 · Accepted Answer · answered Dec 18 '14 at 07:58

Use the fact that $\log $ is monotone increasing function:

$$(a-b)(\log(a)-\log(b))\ge0 \implies \left(\frac{a}{b}\right)^{a-b} \ge 1$$

$$(c-a)(\log(c)-\log(a))\ge0 \implies \left(\frac{b}{c}\right)^{b-c} \ge 1$$

$$(b-c)(\log(b)-\log(c))\ge0 \implies \left(\frac{c}{a}\right)^{c-a} \ge 1$$

Multiply the three inequality yields your desired form !

Prove that $a^ab^bc^c\ge (abc)^{(a+b+c)/{3}}$

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