a b c都为正实数且a+b+c=1求1/(a+b)+1/(b+c)+1/(c+a)大于等于9/2

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abc都为正实数且a+b+c=1求1/(a+b)+1/(b+c)+1/(c+a)大于等于9/2abc都为正实数且a+b+c=1求1/(a+b)+1/(b+c)+1/(c+a)大于等于9/2abc都为正

a b c都为正实数且a+b+c=1求1/(a+b)+1/(b+c)+1/(c+a)大于等于9/2
a b c都为正实数且a+b+c=1求1/(a+b)+1/(b+c)+1/(c+a)大于等于9/2

a b c都为正实数且a+b+c=1求1/(a+b)+1/(b+c)+1/(c+a)大于等于9/2
因为a,b,c>0,由柯西不等式得:
[1/(a+b)+1/(b+c)+1/(c+a)][(a+b)+(b+c)+(c+a)]≥(1+1+1)^2
所以1/(a+b)+1/(b+c)+1/(c+a)≥9/2
当1/(a+b)^2=1/(b+c)^2=1/(c+a)^2时,取到等号,易知a=b=c,联立a+b+c=1,得a=b=c=1/3时,1/(a+b)+1/(b+c)+1/(c+a)≥9/2取到等号