Can TikZ rounded corners make a bulging curvy triangle?

Question

I would like to draw a smoothly rounded figure in TikZ as below, but with simpler (and easier to modify) code. I mean without hand calculating good tangents for the curve at all vertices. That would make it easier to do less symmetric triangles, and other polygons.

Note that unlike the question

Drawing a triangle with rounded corners in TikZ

I want the outline to curve in and out to make bulges.

I wonder if the rounded corners option will let me do this by drawing a hexagon with alternate vertices closer and farther from the center. My attempts so far have failed, for reasons that the TikZ and PGF Manual describes on p. 587. Even when I make the curve of the corners very small the lines ''suddenly extend over the other end" which looks terrible. This may just be more than the option can be expected to do.

But is there a way to do it?

\documentclass{amsart}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
      \draw (0,3.5) 
      to [out=0,in=120]  (1.5,2.7)
      to [out=-60,in=50] (3,0)
      to [out=-130,in=0] (0,.05)
      to [out=180,in=-50] (-3,0)
      to [out=130,in=-120] (-1.5,2.7)
      to [out=60,in=180] (0,3.5);
    \end{tikzpicture}
\end{document}

Answer 1

A simple way (same idea as here: Tikz Macro with list of arguments and subroutine calculation)

\documentclass{amsart}
\usepackage{tikz}

\begin{document}

    \tikzset{
        bulging/.style 2 args={
            every to/.style={bend left=#1},
            preaction={
                  fill,draw,join=round,line width=#2+1},
            white,fill,draw,join=round,line width=#2,
        },
    }

    \tikz\draw[bulging={20}{60}](0,0)to(3,0)to(2,2)to cycle;
    %
    \hfill
    %
    \tikz\draw[bulging={40}{50}](0,0)to(3,0)to(4,2)to(2,5)to(-1,3)tocycle;

    \vfill

    \tikz\draw[bulging={50}{50}](0,0)to(4,0)to(6,2)to(7,4)to(5,8)to(1,8)to(-1,5)tocycle;

\end{document}