<s>[INST] 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 
 [/INST] 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 .
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