Question

设f(x)=lg,其中a∈R.                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                        

如果0＜a≤1,求证:当x≠0时,有2f(x)＜f(2x).

Answer 1

解:设a₁、a₂、a₃都是实数,由均值不等式得?

 

∴2(a₁a₂+a₁a₃+a₂a₃)≤2(a₁²+a₂²+a₃²).①?

∴(a₁+a₂+a₃)²≤3(a₁²+a₂²+a₃²)(当且仅当a₁=a₂=a₃时,取等号).②?

∵x≠0,∴2^x≠1.?

∴由②式知:(1+2^x+4^xa)²＜3(1+2^2x+4^2xa²).③?

∵0＜a≤1,?

∴4^2xa²＜4^2x·a.结合③式即得.?

∴.?

∴2f(x)＜f(2x).