Question

谁能帮我证明下不等式``急` a>1,b>1,求证：a²/(b-1) +b²/(a-1) ≥8

Answer 1

a>1,b>1,求证：a^2/(b-1)+b^2(a-1)≥8

记x=a-1,y=b-1,那么题目变成x>0,y>0,求证(x+1)^2/y+(y+1)^2/x≥8

即(x+1)^2*x+(y+1)^2*y≥8xy

x^3+2x^2+x+y^3+2y^2+y≥8xy

用均值不等式:（a^2+b^2≥2ab)

因为2x^2+2y^2≥4xy x^3+x≥2x^2 y^3+y≥2y^2

所以x^3+2x^2+x+y^3+2y^2+y≥2x^2+2y^2+4xy ≥8xy

证毕