Question

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.

Answer 1

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