Question

不等式的解答

设x>0 , 求证 x^2+(2/x)≥3 怎么变形可以用均值不等式解啊

Answer 1

x^2+(2/x)=x^2-2x+1)+(2x+2/x)-1=(x-1)^2+(2x+2/x)-1 x>0时(x-1)^2>=0 2x+2/x>=4 所以x^2+(2/x)≥3

Answer 2

证明:
x^2+2/x
=x^2+1/x+1/x
>=3*[x^2*(1/x)*(1/x)]^(1/3)
=3
∴x^2+(2/x)≥3