Let a convex quadrilateral ABCD with perimeter 1，d is the maximum of AB,AC,AD,BC,BD,CD,prove that d is not less than 1/3 we can prove that parallelogram ABCD with perimeter 1，than one of AC,BD is more than 1/3 but the general case is very difficult to solve.
The answer given by εδ (a kite inscribed in a Reuleaux triangle) can be found in Ball, D. G. (1973), "A generalisation of π", Mathematical Gazette 57 (402): 298–303, doi:10.2307/3616052, JSTOR 3616052; He doesn't give an explicit proof that this is optimal but says it can be done by "some tedious but not very difficult trigonometry. See also Griffiths, David; Culpin, David (1975), "Pioptimal polygons", Mathematical Gazette 59 (409): 165–175, doi:10.2307/3617699, JSTOR 3617699 for extensions to higher order polygons. 


It is not true. Take $A=0$, $B=1$, $C=e^{\frac\pi6 i}$ $D=e^{\frac\pi3 i}$. Then $d=1$ and the perimeter is $2+\tfrac1{\cos\frac{\pi}{12}} >3$. I am sure that $\frac1{2+\tfrac1{\cos\frac{\pi}{12}}}$ is the optimal bound. 

