Question

已知:a>0,b>0,c>0,求证:(a+b)(a分之1+b分之1)≥4

Answer 1

(a+b)(1/a+1/b)

=1+a/b+b/a+1

=2+(a/b+b/a)

a/b>0,b/a>0

所以a/b+b/a≥2√(a/b*b/a)=2

所以(a+b)(1/a+1/b)≥2+2

即(a+b)(1/a+1/b)≥4