<s>[INST] $a,b,c >0$, and $ab+bc+ca=3$, prove $(a^ab^bc^c)^{\frac{3}{a+b+c}} \geqslant \sqrt[3]{\frac{a^3+b^3+c^3}{3}}$ $a,b,c >0$, and $ab+bc+ca=3$, prove
$$(a^ab^bc^c)^{\frac{3}{a+b+c}} \geqslant \sqrt[3]{\frac{a^3+b^3+c^3}{3}}$$
I think the equality is only achieve when $a=b=c=1$. The condition $ab+bc+ca=3$ is necessary. I used the estimation $x^x \geqslant \frac12 (x^2+1)$ but cannot proceed further. 
 [/INST] It is not an answer, but rather some results which can be helpful for the further investigations. 


*

*$abc \leq 1$
$\rhd$ $$\frac{ab +bc + ac}{3} \geq \sqrt[3]{a^2b^2c^2}\Rightarrow\sqrt[3]{a^2b^2c^2}\leq 1$$
$\lhd$

*One can prove that
$$\forall a_i > 0,\forall p_i>0: \sum\limits_{1}^{n}p_i = 1$$
the following inequality is fulfilled
$$\sum\limits_{i =1}^{n}p_i a_i \geq \prod\limits_{i = 1}^{n}a^{p_i}_{i}.$$
Let us denote
\begin{align} 
\frac{a}{a+b+c} &= \alpha\\
\frac{b}{a+b+c} &= \beta,\\ 
\frac{c}{a+b+c} &= \gamma,
\end{align}
then $\alpha+\beta+\gamma = 1$ and, according to the previous formula ($a_i = \frac1a, \frac1b, \frac1c$), 
$$(a^ab^bc^c)^{\frac{3}{a+b+c}} =(a^\alpha b^\beta c^\gamma)^3 \geq \Big(\frac{a+b+c}{3}\Big)^3 = \Big(\frac{a^3 + b^3 + c^3 + 9(a+b+c) - 3abc}{27}\Big)\geq \Big(\frac{a^3 + b^3 + c^3}{27} + \frac{a+b+c}{3} - \frac19\Big),$$
where the last inequality uses 1.
</s>