Question

帮忙解一道高一数学题 已知a,b,c∈R+,求证:(a^2+1)(b^2+1)(c^2+1)≥8abc

Answer 1

a^2+1≥2√(a^2*1)=2a
同理b^2+1≥2b
c^2+1≥2c
所以原不等式成立
(均值不等式：a+b≥2√ab)

Answer 2

a^2+1≥2a
b^2+1≥2b
c^2+1≥2c
已知a,b,c∈R+
(a^2+1)(b^2+1)(c^2+1)≥8abc

Answer 3

(a-1)^2>=0
=> a^2+1>=2a (1)
For the same reason:
b^2+1>=2b (2)
c^2+1>=2c (3)
Multiplying (1), (2), and (3) together, we can get the above inequality.