Circumference of Convex Shapes

That's a standard inequality, though perhaps not well enough known for the usual "take it to stackexchange" comment. Denote by $\partial \Sigma$ the boundary of any set $\Sigma$ in the plane. In the case that $s$ is a polygon we can use induction on the number $k$ of edges of $\partial s$ not contained in the $\partial S$. If $k=0$ we're done. If $k>0$, choose an edge $e$ of $s$ not in $\partial S$, and let $H$ be the closed half-plane such that $H \supset s$ and $\partial H \supset e$. Then $S' := S \cap H$ is a convex set containing $s$ whose boundary is shorter than $\partial S$ because we've replaced part of $\partial S$ with the line segment joining the same two points. Moreover $\partial s$ has $k-1$ edges not contained in $\partial S'$. This completes the induction step and the proof.

That argument applies more generally when $\partial s \setminus \partial S$ is polygonal. If it's not, we can reduce to that case via a limiting argument, replacing each component of $\partial s \setminus \partial S$ by an arbitrarily close polygonal path.

Your result, including your proposed generalization and the further generalization to surface area in higher dimensions, is an immediate consequence of Cauchy's surface area formula. This states that, up to a constant depending only on the dimension, the surface area of a convex body is the average of the areas of its 1-codimensional projections. I don't know a good reference in a web page, but see for example Klain and Rota's book.

For the triangle problem: Euclid I, 21:

If from the ends of one of the sides of a triangle two straight lines are constructed meeting within the triangle, then the sum of the straight lines so constructed is less than the sum of the remaining two sides of the triangle, but the constructed straight lines contain a greater angle than the angle contained by the remaining two sides.