∵ abc成等比數(shù)列,
∴ b²=a×c；b=a×q；c=a×q²
f(a)+f(c)=[log2^(a+2)]+[log2^(c+2)]
=log2^[ac+2(a+c)+4]
=log2^[b²+2(a+c)+4]
2f(b)=log2^(b+2)²
=log2^(b²+4b+4)
∵ a、b、c、q均大于0
2(a+c)-4b=2a×(1+q²)-4aq
=2a(1+q²-2q)
=2a(1-q)²>0
函數(shù)f(x)在(-2,+∞)上為單調(diào)遞增函數(shù),
∴ f(a)+f(c)>2f(b)