Question

已知x+y+z=3,求证x^4+y^4+z^4≥3

Answer 1

因(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=9
又由基本不等式知x^2+y^2+z^2≥xy+yz+zx
则x^2+y^2+z^2+2(x^2+y^2+z^2)≥9
即x^2+y^2+z^2≥3
于是(x^2+y^2+z^2)^2≥9
即x^4+y^4+z^4+2(x^2y^2+y^2z^2+z^2x^2)≥9
又由基本不等式知x^4+y^4+z^4≥x^2y^2+y^2z^2+z^2x^2
则x^4+y^4+z^4+2(x^4+y^4+z^4)≥9
即x^4+y^4+z^4≥3

Answer 2

即证明2x^4+2y^4+2z^4>=6
即证(x^4+y^4)+(y^4+z^4)+(x^4+z^4)>=6
分别证明x^4+y^4>=2
y^4+z^4>=2
x^4+z^4>=2

追问

怎么证>=2

追答

x+y+z=3>=3(xyz)^(1/3)
即(xyz)^(1/3)3
综上有x^4+y^4+z^4>=3