Question

已知函數(shù)f(x)=lg,0＜a≤1，給定n∈N^*，n≥2.

求證：f(2x)＞2f(x)(x≠0)對(duì)任意n∈N^*且n≥2恒成立.

Answer 1

證明:f(2x)＞2f(x)

［1+2^x+3^x+…+(n-1)^x+a·n^x］²＜n×［1+2^2x+3^2x+…+(n-1)^2x+a·n^2x］,而由柯西不等式得(1·1+1·2^x+1·3^x+…+1·(n-1)^x+a·n^x)²＜(1²+…+1²)×［1+2^2x+3^2x+…+(n-1)^2x+a²·n^2x］(x=0時(shí)取等號(hào))≤n×［1+2^2x+3^2x+…+(n-1)^2x+a·n^2x］(0＜a≤1),結(jié)論得證!