展开左式,欲证结论即:abx1^2+abx2^2+(a^2+b^2)*x1x2≥x1x2即ab(x1^1+x2^2)+(a^2+b^2)*x1x2≥x1x2因x1,x2为正实数,故x1^1+x2^2≥2x1x2那么左式≥ab(2x1x2)+(a^2+b^2)*x1x2=x1x2(a^2+2ab+b^2)=x1x2(a+b)^2=x1x2成立!...