Prove $a^{ab}b+b^{bc}c+c^{ca}a \geqslant \sqrt[6]{5}$

$a,b,c >0$, and $a+b+c=3$, prove $$ a^{ab}b+b^{bc}c+c^{ca}a \geqslant \sqrt[6]{5}$$

I try to substitute $c=3-a-b$ to reduce the number of variables, but cannot further proceed to solve the problem. I made an Excel spreadsheet and test 100 pairs of $(a,b,c)$, it seems that the inequality is correct. I cannot even find where the equality occurs. Please help. This is a very unconventional problem

Hint: for $x>0$ and $y>0$ and $z>0$ we have : $$x+y+z\geq \left(\frac{x+1}{y+1}+\frac{y+1}{x+1}+\frac{x+1}{z+1}+\frac{z+1}{x+1}+\frac{z+1}{y+1}+\frac{y+1}{z+1}-1\right)^{\frac{1}{6}}\geq (5)^{\frac{1}{6}}$$ Method: If we put $x+y+z=\lambda$

with $x=a^{ab}a$

$y=b^{bc}b$

$z=c^{ac}c$

And $a+b+c=3$

The question is : when is the maximum reached in the following expression? $$\frac{x+1}{y+1}+\frac{y+1}{x+1}+\frac{x+1}{z+1}+\frac{z+1}{x+1}+\frac{z+1}{y+1}+\frac{y+1}{z+1}-1$$ The maximum is reached when $x=y=0$ and $z=\lambda$

Following this we have this inequality : $$\lambda\geq \left(1+2(\lambda+1+\frac{1}{1+\lambda})\right)^{1/6}$$ It occurs for $\lambda\simeq 1.36897$

Now the idea is to repeat the same reasoning with the following expression : $$\frac{x+2}{y+2}+\frac{y+2}{x+2}+\frac{x+2}{z+2}+\frac{z+2}{x+2}+\frac{z+2}{y+2}+\frac{y+2}{z+2}-1$$ We obtain this : $$\lambda\geq \left(1+2(\frac{2+\lambda}{2}+\frac{2}{2+\lambda})\right)^{\frac{1}{6}}$$ Its occurs for $\lambda\simeq 1.32985$

Now the idea is to take the following expression :

$$\frac{x+n}{y+n}+\frac{y+n}{x+n}+\frac{x+n}{z+n}+\frac{z+n}{x+n}+\frac{z+n}{y+n}+\frac{y+n}{z+n}-1$$ And with the same reasoning we obtain :

$$\lambda\geq \left(1+2(\frac{n+\lambda}{n}+\frac{n}{n+\lambda})\right)^{\frac{1}{6}}$$

We take the limit :

$$\lambda\geq \lim\limits_{n \to \infty}\left(1+2(\frac{n+\lambda}{n}+\frac{n}{n+\lambda})\right)^{\frac{1}{6}}$$ And we obtain $$\lambda \geq (5)^{\frac{1}{6}}$$

If I'm wrong tell me quickly .