Inequality $\Big(\frac{1}{3}\Big)^{\frac{1}{27}}\leq ab^{abc}+bc^{abc}+ca^{abc}\leq 1$

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It's a problem from my own :

Let $a,b,c>0$ such that $a+b+c=1$ then we have : $$\Big(\frac{1}{3}\Big)^{\frac{1}{27}}\leq a\Big(b^{abc}\Big)+b\Big(c^{abc}\Big)+c\Big(a^{abc}\Big)=f(a,b,c)\leq 1$$

The idea was to create an inequality where $f(a,b,c)$ takes his values in a very little interval because we have $\Big(\frac{1}{3}\Big)^{\frac{1}{27}}-1<-0.0398$

Furthermore I create this where all the classical method fails .I think to Am-Gm , Jensen , Tchebytchev and so on ...

My first idea was to use rearrangement inequality but it doesn't works too .

Thirdly my idea was to use the Lagrange multiplier but I'm skeptical about this method .

An other point of view is to put $abc=\alpha=constant$ and study the behavior of $f(a,b,c)$.

That's all for me if you have a hint it would be great .

Thanks a lot for sharing your time and knowledge .

Edit

As pointed out by WE tutorial school The RHS is obvious but what happend for the LHS ?