分类：教育/科学-理工学科-数学
问题：设正实数a、b、c满足1/(a+b+1)+1/(b+c+1)+1? 
证明:a+b+c>=ab+bc+ca.
回答：证明:由auchy不等式得(a+b+1)(a+b+c^2)>=(a+b+c)^2--->1/(a+b+c)=<(c^2+a+b)/(a+b+c)^2 ......(1)同理,1/(b+c+1)=<(a^2+b+c)/(a+b+c)^2 ......(2)1/(c+a+1)=<(b^2+c+a)/(a+b+c)^2由(1)+(2)+(3),得1/(a+b+1)+1/(b+c+1)+1/(c+a+1)=<[a^2+b^2+c^2+2(a+b+c)]/(a+b+c)^2而已知1/(a+b+1)+1/(b+c+1)+1/(c+a+1)>=1故[a^2+b^2+c^2+2(a+b+c)]/(a+b+c)^2=<1--->a^2+b^2+c^2+2(a+b+c)>=(a+b+c)^2--->a^2+b^2+c^2+2a+2b+2c>=a^2+b^2+c^2+2ab+2bc+2ca--->a+b+c>=ab+bc+ca证毕.