① Let the three sides of △ ABC be a, B and C respectively, and try to prove that: a < 12 (a + B + C). ② let the four sides of a quadrilateral be a, B, C and D in turn, and the two diagonals be e and f respectively, and prove that: e + F > 12 (a + B + C + D)

① Let the three sides of △ ABC be a, B and C respectively, and try to prove that: a < 12 (a + B + C). ② let the four sides of a quadrilateral be a, B, C and D in turn, and the two diagonals be e and f respectively, and prove that: e + F > 12 (a + B + C + D)

① It is proved that: ∵ B + C ＞ a, ∵ 12b + 12C ＞ 12a, ∵ 12b + 12C + 12a ＞ 12a + 12a, ∵ 12 (a + B + C) ＞ a, that is, a ＜ 12 (a + B + C); ② it is proved that: obviously n + X ＞ a, x + m ＞ B, y + m ＞ C, N + y ＞ D, so: 2 (x + y + m + n) ＞ a + B + C + D, that is: 2 (E + F) ＞ a + B + C + D, so: e + F