The answer is 5.16.

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Taking logarithm and equating both of functions:

$\ln{x^2}=\ln{x^{\frac{1}{x}}}\\\left(2-\dfrac{1}{x}\right)\ln{x}=0\implies{x=1~~\text{and}~~x=\dfrac{1}{2}}$

So, we have $A\left(1, 1\right)~~\text{and}~~B\left(\dfrac{1}{2}, \dfrac{1}{4}\right)$ .

Now, taking the derivative of the second function,

$\ln{y}=\dfrac{1}{x}\ln{x}\\\dfrac{y'}{y}=-\dfrac{1}{x^2}\ln{x}+\dfrac{1}{x}\times{\dfrac{1}{x}}=\dfrac{1}{x^2}\left(1-\ln{x}\right)\\\dfrac{\sqrt[x]{x}}{x^2}\left(1-\ln{x}\right)=0\implies{x=\large{e}=p}$

and we have $C\left(\large{e}, 0\right)$ . Now, the distance of point $A$ to $B$ , $A$ to $C$ , and $B$ to $C$ are $0.901$ , $1.988$ , and $2.232$ respectively, and, finally the perimeter of triangle ABC is $5.12$