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
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