>>145
　(aab+bbc+cca +1) - (aa+bb+cc) = 1 - aa(1-b) - bb(1-c) - cc(1-a)
　≧ 1 - a(1-b) - b(1-c) - c(1-a)
　= (1-a)(1-b)(1-c) + abc
　≧ 0,

>>146
そりゃ、KoMaL
C.1552
　Prove that if 0<a<1 and 0<b<1 then log_a{2ab/(a+b)}・log_b{2ab/(a+b)} ≧ 1,

　A = log(a) < 0,　B = log(b) < 0,
　2ab/(a+b) ≦ √(ab),
　log_a{2ab/(a+b)} = log{2ab/(a+b)}/ log(a) ≧ log(ab)/{2log(a)} = (A+B)/(2A) > 0,
　log_b{2ab/(a+b)} ≧ (A+B)/(2B) > 0,
辺々掛けて
　(左辺) ≧ (A+B)^2 /(4AB) ≧ 1,