ABC is a triangle. Prove that AB+AC+AD+AE < R

[nguoidep_68]

Choose five points A;B;C;D and E on a sphere with radiusR such that [tex]\angle{BAC}=\angle{CAD}=\angle{DAE}=\angle{EAB}=\frac{2}{3}.\angle{BAD}=\frac{2}{3}.\angle{CAE}[/tex]Prove that [tex]AB+AC+AD+AE \leq 4\sqrt{2}R[/tex]