已知x、y、z、是正实数,且x+y+z=xyz,求1/(x+y)+1/(y+z)+1/(x+z)的最大值.
人气:375 ℃ 时间:2019-09-23 09:30:34
解答
配凑柯西不等式1/(x+y)+1/(y+z)+1/(z+x)≤[1/2(xy)^0.5]+[1/2(yz)^0.5]+[1/2(zx)^0.5]=(1/2){1*[z/(x+y+z)]^0.5+1*[x/(x+y+z)]^0.5+1*[y/(x+y+z)]^0.5}≤(1^2+1^2+1^2)[x/(x+y+z)+y/(x+y+z)+z/(x+y+z)]^0.5=√3/2(这种证法综合运用了柯西不等式和基本不等式) 因此λ只要大于等于√3/2就行了
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