若a.b∈R+,且3a+b=1,则P=1/a + 1/b的小值是
热心网友
因为3a+b=1,所以:1/a+1/b=(1/a+1/b)×1=(1/a+1/b)(3a+b)=4+3a/b+b/a,而又因为a,b∈R+,所以3a/b+b/a≥2√[(3a/b)×(b/a)]=2√3所以1/a+1/b=4+3a/b+b/a≥4+2√3,所以1/a+1/b的最小值为4+2√3
若a.b∈R+,且3a+b=1,则P=1/a + 1/b的小值是
因为3a+b=1,所以:1/a+1/b=(1/a+1/b)×1=(1/a+1/b)(3a+b)=4+3a/b+b/a,而又因为a,b∈R+,所以3a/b+b/a≥2√[(3a/b)×(b/a)]=2√3所以1/a+1/b=4+3a/b+b/a≥4+2√3,所以1/a+1/b的最小值为4+2√3