Question

(1)已知正數(shù)a、b、c成等比數(shù)列，求證:a²-b²+c²≥(a-b+c)²;

(2)設(shè)a、b∈R,求證:a²+b²≥2(a-b-1).

Answer 1

思路分析:證明不等式，通常可以看作是比較兩式大小問題.

(1)證明:∵ac=b²,b＞0,∴b=.

∴a²-b²+c²-(a-b+c)²=a²-b²+c²-a²-b²-c²+2ab-2ac+2bc

=2ab-2b²-2ac+2bc=2ab-4b²+2bc

=2b(a-2b+c)=2b()²≥0.

∴a²-b²+c²≥(a-b+c)².

(2)證明:∵a²+b²-2(a-b-1)=a²+b²-2a+2b+2

=a²-2a+1+b²+2b+1

=(a-1)²+(b+1)²≥0,

∴a²+b²≥2(a-b-1).