设直线方程为:x/a+y/b=1,该直线经过点(4,1),所以4/a+1/b=1下面给出三个思路:1.利用柯西不等式得到a+b=(a+b)(4/a+1/b)≥[√a*√(4/a)+√b*√(1/b)]^2=[2+1]^2=9,当a=4/a,b=1/b时取等号即a=2,b=1时等号成立。2.1/b=1-4/a=(a-4)/a,b=a/(a-4)=1+4/(a-4)a+b=a+1+4/(a-4)=a-4+4/(a-4)+5≥2√[(a-4)4/(a-4)]+5=2*2+5=9.当a-4=4/(a-4),即a-4=2或-2,a=6或a=2,(此时可得b<0,舍去)3.设a+b=m,b=m-a,4/a+1/(m-a)=1...
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