问题补充:
已知a、b、c为正数,n是正整数,且,求证:2f(n)≤f(2n).
答案:
证明:∵a2+b2≥2ab
∴(an+bn+cn)2
=a2n+b2n+c2n+2an?bn+2an?cn+2bn?cn
≤3(a2n+b2n+c2n)
∴lg(an+bn+cn)2≤lg[3(a2n+b2n+c2n)]
∴lg(an+bn+cn)2≤lg(a2n+b2n+c2n)+lg3
∴2lg(an+bn+cn)≤lg(a2n+b2n+c2n)+lg3
∴2[lg(an+bn+cn)-lg3]≤lg(a2n+b2n+c2n)-lg3
∴2f(n)≤f(2n)
解析分析:由基本不等式的推论a2+b2≥2ab,可得(an+bn+cn)2=a2n+b2n+c2n+2an?bn+2an?cn+2bn?cn≤3(a2n+b2n+c2n),进而根据对数的运算性质及,可证得结论.
点评:本题考查的知识点是对数函数的单调性,其中根据基本不等式的推论得到(an+bn+cn)2=a2n+b2n+c2n+2an?bn+2an?cn+2bn?cn≤3(a2n+b2n+c2n),是解答本题的关键.