∵a+b=1平方得
a²+b²+2ab=1.
∴(1/a²)+(1/b²)=[(a²+b²+2ab)/a²]+[(a²+b²+2ab)/b²]
=2+[(a²/b²)+(b²/a²)]+2[(b/a)+(a/b)]
根据均值不等式:
(a²/b²)+(b²/a²)≥2
(b/a)+(a/b)≥2
∴(1/a²)+(1/b²)≥2+2+2×2=8.
∵a+b=1平方得
a²+b²+2ab=1.
∴(1/a²)+(1/b²)=[(a²+b²+2ab)/a²]+[(a²+b²+2ab)/b²]
=2+[(a²/b²)+(b²/a²)]+2[(b/a)+(a/b)]
根据均值不等式:
(a²/b²)+(b²/a²)≥2
(b/a)+(a/b)≥2
∴(1/a²)+(1/b²)≥2+2+2×2=8.