Prove that (sinA+sinB)(sinB+sinC)(sinC+sin A) > sin A sinB sinC in a triangle ABC

we have in a triangle a + b > c

so sin A + sin B > sin C ..1

as a/ sin A = b/ sin B = c/ sin C

similarly

( sin B + sin C) > sin A ..2

sin C + sin A > sin B ...3

by multiplying (1) (2) and (3) and knowing that both sides are positive ( sin is positive for any angle of a triangle)

we get the result

(sinA+sinB)(sinB+sinC)(sinC+sinA)>sinA sinB sinC

proved