已知a,b,c为正实数,且ab+bc+ca=1,证明1/(a^2+1)+1/(b^2+1)+1/(c^2+1)=3/4是怎么来的_数学解答

已知a,b,c为正实数,且ab+bc+ca=1,证明1/(a^2+1)+1/(b^2+1)+1/(c^2+1)<=9/4
请问这步解析：原不等式等价于a^2/(a^2+1)+b^2/(b^2+1)+c^2/(c^2+1)>=3/4是怎么来的

人气：300 ℃　时间：2020-05-13 04:31:17

解答

就是两边同时被3减去
3-[1/(a^2+1)+1/(b^2+1)+1/(c^2+1)]
=[1-1/(a^2+1)]+[1-1/(b^2+1)]+[1-1/(c^2+1)]
=a^2/(a^2+1)+b^2/(b^2+1)+c^2/(c^2+1)