xlnx+ylny-xln(x+y)-yln(x+y)-(x+y)ln(1/2)=xln[x/(x+y)]+yln[y/(x+y)]-(x+y)ln(1/2)=-xln(1+y/x)-yln(1+x/y)-(x+y)ln(1/2)=-x[ln(1+y/x)+y/xln(1+z/y)-(1+y/x)ln2]令y/x=t即证ln(1+t)+tln(1+1/t)-(1+t)ln2<0由于x、y地位对等,所以设y>x,即t>1 构造f(t)=ln(1+t)+tln(1+1/t)-(1+t)ln2求导f"(t)=ln(1+t)-lnt-ln2=ln(1+1/t)-ln2<0所以f(t)是减函数f(t)<f(1)=0所以ln(1+t)+tln(1+1/t)-(1+t)ln2<0所以-x[ln(1+y/x)+y/xln(1+z/y)-(1+y/x)ln2]>0即xlnx+ylny>(x+y)ln[(x+y)/2]>0求一阶导不算犯规吧?要纯构造也行,不过非常麻烦。 不好意思,疏忽了,那儿是f(1)