证明：(1)由 |lga| = |lgb| ,得 lga = lgb 或 lga = -lgb
所以 a = b 或 a = 1/b
因为 0 1 , 所以 b² > 1 , 所以 1/b² ∈ (0, 1)
又 2|lg[(a+b)/2] = |lgb| , b > 1 , a = 1/b
则 2|lg[(1/b + b)/2]| = lgb
由基本不等式得 (1/b + b)/2 > 2/2 = 1
则 2lg[(1/b + b)/2] = lgb
即 [(1/b + b)/2]² = b
即 1/b² + 2 + b² = 4b
则 2< 4b - b² = 2 + 1/b²