WLOG $ a\geq b\geq c$, then $ \ln{a}\geq \ln{b}\geq \ln{c}$ . Because these two are non decreasing sequences by Chebyschev we have: $ a\ln{a} + b\ln{b} + c\ln{c} \geq \frac{(a+b+c)(\ln{a} + \ln{b} + \ln{c})}{3}$ By the properties of logarithms this is equivalent to: $ \ln{a^{a}b^{b}c^{c}}\geq \ln{abc^{\frac{a+b+c}{3}}}$ And because the logarithms are bijective we are done.

By the weighted GM-HM inequality on $ a$ with weight $ a$, $ b$ with weight $ b$ and $ c$ with weight $ c$: Then $ (a^{a}b^{b}c^{c})^{\frac{1}{a+b+c}}\ge\frac{a+b+c}{\frac{a}{a}+\frac{b}{b}+\frac{c}{c}}=\frac{a+b+c}{3}$ Finally using the AM-GM inequality on $ a,b,c$: $ \frac{a+b+c}{3}\ge (abc)^{\frac{1}{3}}$ Combining the two gives: $ (a^{a}b^{b}c^{c})^{\frac{1}{a+b+c}}\ge\frac{a+b+c}{3}\ge (abc)^{\frac{1}{3}}$ Hence $ a^{a}b^{b}c^{c}\ge (abc)^{\frac{a+b+c}{3}}$ as required.

First define that $ a \ge b \ge c$. Now note that $ a^{\frac{a-b}{2}} \ge b^{\frac{a-b}{2}}$ $ a^{a-\frac{a+b}{2}}b^{b-\frac{a+b}{2}} \ge 1$ $ a^a b^b \ge (ab)^{\frac{a+b}{2}}$ Multiplying the above for all pairings of $ a,b,c$, gives that $ (a^a b^b c^c)^2 \ge (abc)^{\frac{a+b+c}{2}} \cdot (a^a b^b c^c)^{\frac{1}{2}}$ $ (a^a b^b c^c)^{\frac{3}{2}} \ge (abc)^{\frac{a+b+c}{2}}$ $ a^{a}b^{b}c^{c}\ge (abc)^{(a+b+c)/3}$ As desired.

Also posted here.

As $a$,$b$ and $c\in \mathbb{R^+}$ and $a$,$b$ and $c>0$,by the AM-GM inequality ,\[a+b+c\ge 3(abc)^\frac{1}{3}\]. Taking logarithm on both sides, we have \[\log \frac{a+b+c}{3}\ge \frac{1}{3} (\log a+ \log b+\log c)\dots \boxed{1}\] Now consider the function $f(x)=x\log x$.$f''>0$, i.e. $f$ is convex.So by Jensen's inequality, we get \[\frac{1}{3}(a \log a+b\ log b+c \log c)\ge \frac{a+b+c}{3}\log (\frac{a+b+c}{3})\] \[\ge \frac{1}{3} (\log a+ \log b+\log c)\cdot \frac{a+b+c}{3}\](using $\boxed{1}$) which gives us the desired inequality on simplification. $\blacksquare$

Solution without any prior knowledge to any inequality formulas: Note that $(a^ab^bc^c)(a^bb^cc^a)(a^cb^ac^b)=\left((abc)^{(a+b+c)/3}\right)^3$. So if we can prove that $a^ab^bc^c\ge a^bb^cc^a$ and $a^ab^bc^c\ge a^cb^ac^b$, then we are done. WLOG let $a\ge b\ge c$. Note that $(a^ab^bc^c)\cdot \left(\dfrac{c}{a}\right)^{a-b}\cdot \left(\dfrac{c}{b}\right)^{b-c}=a^bb^cc^a$. Since $\dfrac{c}{a} \le 1$, $\dfrac{c}{b} \le 1$, $a-b \ge 0$, and $b-c \ge 0$, it follows that $a^ab^bc^c \ge a^bb^cc^a$. Note that $(a^ab^bc^c)\cdot \left(\dfrac{b}{a}\right)^{a-b}\cdot \left(\dfrac{c}{a}\right)^{b-c}=a^cb^ac^b$. Since $\dfrac{b}{a} \le 1$, $\dfrac{c}{a} \le 1$, $a-b \ge 0$, and $b-c \ge 0$, it follows that $a^ab^bc^c \ge a^cb^ac^b$. Thus, $(a^ab^bc^c)^3\ge (a^ab^bc^c)(a^bb^cc^a)(a^cb^ac^b)=\left((abc)^{(a+b+c)/3}\right)^3$, and cube-rooting both sides gives $a^ab^bc^c\ge (abc)^{(a+b+c)/3}$ as desired.

WLOG $a \leq b \leq c$ so $\log_{abc}(a) \leq \log_{abc}(b) \leq \log_{abc}(c)$. Then, $\log_{abc}(a) \leq \frac{1}{3}$ and $\log_{abc}(c) \geq \frac{1}{3}$. Now we look take the log base abc of the inequality giving $a + (b-a) \log_{abc}(bc) + (c-b) \log_{abc}(c) \geq \frac{a+b+c}{3} $. Multiplying by 3, gives $3a + 3(b-a) \log_{abc}(bc) + 3(c-b) \log_{abc}(c) \geq a+b+c $. We can cancel the a's giving $2a + 3(b-a)\log_{abc}(bc) + 3(c-b)\log_{abc}(c) \geq b+c$. We know $\log_{abc}(bc) \geq \frac{2}{3}$ since $\log_{abc}(a) \leq \frac{1}{3}$. Then, if the inequality holds when we substitute $\log_{abc}(bc)$ with $\frac{2}{3}$, it holds for all cases. Also, $\log_{abc}(c) \geq \frac{1}{3}$ so if the inequality holds when we substitute $\log_{abc}(c)$ with $\frac{1}{3}$ then the inequality will hold for all cases. Substituting these values gives, $2a+3(b-a)\frac{2}{3} + 3(c-b)\frac{1}{3} \geq b+c$ and simplifying gives $b+c \geq b+c$ and since this is true, the inequality holds since all steps are reversible.

I'm essentially proving Chebyshev with smoothing here, but I didn't know what Chebyshev was before, so it's fine. We want to show \[\sum_{\mathrm{cyc}}a\log a\ge\frac{a+b+c}{3}\sum_{\mathrm{cyc}}\log a.\]We'll first prove the two variable version. It rearranges to $\frac{a-b}{2}\log a\ge\frac{a-b}{2}\log b$ which is clearly true, so WLOG we may assume $a=b$. The inequality we WTS then reduces to $\frac{2a-2b}{3}\log a\ge\frac{2a-2b}{3}\log b$ which is also true, so we're done.

Since the inequality is homogeneous we can assume $a+b+c=1$.Then by the weighted GM-HM inequality and AM-GM inequality: $$a^{a} b^{b} c^{c} \ge \frac{a+b+c}{\frac{a}{a}+\frac{b}{b}+\frac{c}{c}} = \frac{a+b+c}{3} \ge (abc)^\frac{1}{3} = (abc)^\frac{a+b+c}{3}$$as desired $\blacksquare$

We will transworm this inequality: $a^ab^bc^c \geq \sqrt[3]{abc}^{a+b+c}$. Now: $a^{\frac{a}{\sqrt[3]{abc}}} \cdot b^{\frac{b}{\sqrt[3]{abc}}} \cdot c^{\frac{c}{\sqrt[3]{abc}}} \geq (\sqrt[3]{abc})^{\frac{a}{\sqrt[3]{abc}}} \cdot (\sqrt[3]{abc})^{\frac{b}{\sqrt[3]{abc}}} \cdot (\sqrt[3]{abc})^{\frac{c}{\sqrt[3]{abc}}}$. And now, we can simiplify this to get: $(\frac{a}{\sqrt[3]{abc}})^{\frac{a}{\sqrt[3]{abc}}} \cdot (\frac{b}{\sqrt[3]{abc}})^{\frac{b}{\sqrt[3]{abc}}} \cdot (\frac{c}{\sqrt[3]{abc}})^{\frac{c}{\sqrt[3]{abc}}} \geq 1$. We will now make a substitution. Let $m=(\frac{a}{\sqrt[3]{abc}})^{\frac{a}{\sqrt[3]{abc}}}, n=(\frac{b}{\sqrt[3]{abc}})^{\frac{b}{\sqrt[3]{abc}}}, p=(\frac{c}{\sqrt[3]{abc}})^{\frac{c}{\sqrt[3]{abc}}}$. So, $mnp=1$ and we need to prove that $m^mn^np^p \geq 1$ Let $m \geq n \geq p, m \geq 1$ and $mn=\frac{1}{p} \geq 1$ and now we will factorize: $m^mn^np^p=\frac{m^mm^nn^nm^pn^pp^p}{m^nm^pn^p}=m^{m-n}(\frac{1}{p})^{n-p} \geq 1$ $\implies a^ab^bc^c \geq (abc)^{\frac{a+b+c}{3}}$.

We first put all variables to the LHS, so we need to prove that $$1\ge a^\frac{-2a+b+c}{3}b^\frac{a-2b+c}{3}c^\frac{a+b-2c}{3}$$we can ignore the 3 in the denominator and replace it by $2(a+b+c)$ just to make the weights add up to 1 and using Weighted AM-GM, $$1=\frac{1}{2(a+b+c)}\left((a)(\frac{b}{a}+\frac{c}{a})+(b)(\frac{a}{b}+\frac{c}{a})+(c)(\frac{a}{c}+\frac{b}{c})\right)\ge \left((\frac{b}{a})^a(\frac{c}{a})^a(\frac{a}{b})^b(\frac{c}{b})^b(\frac{a}{c})^c(\frac{b}{c})^c\right)^\frac{1}{2(a+b+c)}.$$Thus, the desired result follows.

Lemma : If $x,y>0$, then $x^xy^y\ge x^yy^x$. Proof WLOG, we assume that $x\ge y$, then $x^{x-y}\ge y^{x-y}$. Hence : $\dfrac{x^x}{x^y}\ge\dfrac{y^x}{y^y}$. Therfore : $x^xy^y\ge x^yy^x$. Now applying this lemma we get : $$a^ab^b\ge a^bb^a\qquad (1)$$$${\color{blue}a^ac^c\ge a^cc^a\qquad (2)}$$$${\color{red}b^bc^c\ge b^cc^b\qquad (3)}$$and: $${\color{green}a^ab^bc^c\ge a^ab^bc^c\qquad (4)}$$Now we multiply these inequalities term by term , we obtain : $$a^ab^b{\color{blue}a^ac^c}{\color{red}b^bc^c}{\color{green}a^ab^bc^c}\ge a^bb^a{\color{blue}a^cc^a}{\color{red}b^cc^b}{\color{green}a^ab^bc^c}.$$Thus : $$a^{3a}b^{3b}c^{3c}\ge a^{a+b+c}b^{a+b+c}c^{a+b+c}.$$Therefore : $$\left(a^ab^bc^c\right)^3\ge\left(abc\right)^{a+b+c}.$$

Brut3Forc3 wrote: Prove that if $ a,b,$ and $ c$ are positive real numbers, then \[ a^ab^bc^c \ge (abc)^{(a+b+c)/3}.\] Using GM-HM inequality involving weights followed by usual AM-GM inequality we get $$\left(a^{a}b^{b}c^{c} \right)^{a+b+c} \ge \frac{a+b+c}{3} \ge (abc)^{1/3} \Leftrightarrow \left(a^{a}b^{b}c^{c} \right) \ge (abc)^{a+b+c/3}$$

As a sidenote, instead of WLOGging this problem, we can do the following: Lemma: If $x,y>0,$ then $x^xy^y \ge x^yy^x.$ Proof: Rearrange to get the two identical inqeualities $$\left(\frac{x}{y}\right)^x\ge \left(\frac{x}{y}\right)^y$$and $$\left(\dfrac{y}{x}\right)^y\ge \left(\dfrac{y}{x}\right)^x.$$If $x\ge y,$ then $\frac{x}{y} \ge 1$ and the first inequality is proven. If $y \ge x,$ then $\frac{y}{x} \ge 1$ and the second inequality is proven. Proceed by @2above

ouldyoubba wrote: Lemma : If $x,y>0$, then $x^xy^y\ge x^yy^x$. Proof WLOG, we assume that $x\ge y$, then $x^{x-y}\ge y^{x-y}$. Hence : $\dfrac{x^x}{x^y}\ge\dfrac{y^x}{y^y}$. Therfore : $x^xy^y\ge x^yy^x$. Now applying this lemma we get : $$a^ab^b\ge a^bb^a\qquad (1)$$$${\color{blue}a^ac^c\ge a^cc^a\qquad (2)}$$$${\color{red}b^bc^c\ge b^cc^b\qquad (3)}$$and: $${\color{green}a^ab^bc^c\ge a^ab^bc^c\qquad (4)}$$Now we multiply these inequalities term by term , we obtain : $$a^ab^b{\color{blue}a^ac^c}{\color{red}b^bc^c}{\color{green}a^ab^bc^c}\ge a^bb^a{\color{blue}a^cc^a}{\color{red}b^cc^b}{\color{green}a^ab^bc^c}.$$Thus : $$a^{3a}b^{3b}c^{3c}\ge a^{a+b+c}b^{a+b+c}c^{a+b+c}.$$Therefore : $$\left(a^ab^bc^c\right)^3\ge\left(abc\right)^{a+b+c}.$$ Nice solution! I did the same thing. The original problem is similar to a homogenized inequality, so motivations would definitely be multiplying similar inequalities, especially using $a^ab^b\geq{a^bb^a}$, which is a huge motivator, since a^a appears on one side and a^b appears partially on the other.