I conjecture $3(a^ab^bc^c)^{\frac{1}{a+b+c}}\ge (a^ab^b)^{\frac{1}{a+b}}+(b^bc^c)^{\frac{1}{b+c}}+(c^ca^a)^{\frac{1}{c+a}}$

Question

I Conjecture, for $a, b, c > 0$, $$3\left(a^ab^bc^c\right)^{\dfrac{1}{a+b+c}}\ge \left(a^ab^b\right)^{\frac{1}{a+b}}+\left(b^bc^c\right)^{\frac{1}{b+c}}+\left(c^ca^a\right)^{\frac{1}{c+a}}.$$

This conjecture is based on the problem (use Jensen inequality can prove it):
Prove $a^ab^bc^c\ge (abc)^{\frac{a+b+c}3}$ for positive numbers.

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

It seem is right,I can't find count-example so far,because $(a,b,c)=(1,1,1),(1,2,3)$ is hold,and $(a,b,c)=(x,y,\to 0)$ also hold.and jensen inequality Can't use?