Length of unit circle Let $\it{l} $ be the length of the unit circumference $\{(x,y):||(x,y)||=1\}$ in an arbitrary norm $||\cdot||$ in $\mathbb{R}^2.$ How to prove or disprove the inequalities $\it{l} \ge 6,\, \it{l} \le 8$? I find $\it{l} =8 $ if $||(x,y)||:=|x|+|y|.$ 
 A: Both inequalities are true; they are due to Gołab (1932), see references in section 7.4 of The geometry of Minkowski spaces I by Horst Martini, Konrad J Swanepoel, Gunter Weiss. Moreover, $8$ is attained only when the unit circle is a parallelogram, and 6 is attained only when it's a linear image of a regular hexagon. Two book references for these results are


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*"Minkowski Geometry" by A. G. Thompson, page 113-114

*"Geometry of spheres in normed spaces" by J. J. Schäffer, pages 17-19


Since the complete proof takes a while, I'll only summarize the main steps here. 
$l\ge 6$
Suppose that $C$ is the unit circle for some norm. Fix a point $p\in C$. Draw the   circle $\{x\colon \|x-p\|=1\}$; it will cross $C$ at some point $q$. The points $p,q,q-p, -p, -q, p-q$ are vertices of a hexagon inscribed in $C$, because each of them has a unit norm. Since every side of this hexagon has length $1$, the length of $C$ is at least $6$. (Recall that length is the supremum of lengths of inscribed polygonal curves.)
$l\le 8$


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*There exist two unit vectors $u,v$ such that the parallelogram with vertices $\pm u\pm v$ contains the unit disk (takes some work). 

*The aforementioned parallelogram has length $8$ (clear).

*When one convex domain contains another, the smaller domain has smaller perimeter (easy to prove in Euclidean plane, takes more work in general normed planes).

