a,b属于R ,a b=1,证明(a 1/a)(b 1/b)>=25/4a,b属于R+,a+b=1,证明(a+1/a)(b+1/b)>=25/4
热心网友
证明:∵a,b属于R+,a+b=1∴(a+1/a)(b+1/b)=［a+（a+b）/a］［b+（a+b）/b］=(a+b/a+1)(b+a/b+1)=ab+(a^2/b+b^2/a)+(b/a+a/b)+3≥ab+2√(ab)+2+3=1/4+1+5=25/4(当且仅当b/a=a/b,a^2/b=b^2/a时取等号,此时a=b=1/2,ab=1/4,2√(ab)=1)∴(a+1/a)(b+1/b)≥25/4
热心网友
证明:∵a,b属于R+,a+b=1(a+1/a)(b+1/b)=ab+a/b+b/a+1/ab=(ab+1/ab)+(a/b+b/a)令m=ab+1/ab,n=a/b+b/a,则n=a/b+b/a≥2, ①ab≤[(a+b)/2]^2=1/4,∴0
热心网友
证明:(a+1/a)(b+1/b)=ab+(a/b+b/a)+1/ab=ab+(a/b+b/a)+1/16ab+15/16ab=(ab+1/16ab)+(a/b+b/a)+15/16ab=(ab+1/16ab)+(a/b+b/a)+15(a+b)^/16ab[注:a+b=1]=(ab+1/16ab)+(a/b+b/a)+(15/16)(2+a/b+b/a)≥[2√(ab)(1/16ab)]+[2√(a/b)(b/a)]+(15/16)[2+2√(a/b)(b/a)]=1/2+2+(15/4)=25/4∴(a+1/a)(b+1/b)≥25/4