Can you prove this inequality w i t h o u t use of vectors?

this is Pedoe's inequality:


for two triangles abc and a'b'c' ,with areas F and F' respectively you have :


a^2.((b'^)2 + (c')^2 - (a')^2) + b^2.((a')^2 + (b')^2 - (c')^2) + c^2.((a')^2 + (b')^2 - (c')^2) %26gt;= 16.F.F'


equality stands for similar triangles.

Can you prove this inequality w i t h o u t use of vectors?
Let s=1/2 . (a+b+c), s'=1/2 . (a'+b'+c'),


F=sqrt[s . (s-a) . (s-b) . (s-c)],


F'=sqrt[s' . (s'-a') . (s'-b') . (s'-c')],


16.F.F' = sqrt{[(a+b)^2-c^2] . [c^2-(a-b)^2] . [(a'+b')^2-c'^2] . [c'^2-(a'-b')^2]} %26lt;=


1/2 . {[(a+b)^2-c^2] . [c'^2-(a'-b')^2] + [c^2-(a-b)^2] . [(a'+b')^2-c'^2]} = a^2.((b'^)2 + (c')^2 - (a')^2) + b^2.((a')^2 + (b')^2 - (c')^2) + c^2.((a')^2 + (b')^2 - (c')^2),


when a'/a=b'/b=c'/c, we get the equality.