Question

(x+y)2+(y+x)2

Answer 1

2(x^3 +y^3 +z^3 )-〔返胡岁x^2 (y +z)+ y^2 (x +z) +z^2 (x +y)〕
=(x^3 +y^3+x^3+z^3+y^3+z^3)-(x^2*y+x^2*z+x*y^2+y^2*z+x*z^2+y*z^2〕
=(x^3+y^3-x^2*y-x*y^2)+(x^3+z^3-x^2*z-x*z^2)+(y^3+z^3-y^2*z-y*z^2)
=(x^2-y^2)(x-y)+(x^2-z^2)(x-z)+(y^2-z^2)(y-z)
=(x+y)(x-y)^2+(x+z)(x-z)^2+(y+z)(y-z)^2
因为x,y,z为正数，即x>0,y>0,z>0,
则x+y>0,x+z>0,y+z>0,
又因为
(x-y)^2>=0,(x-z)^2>=0,(y-z)^2>漏睁=0,
所以做橘
2(x^3 +y^3 +z^3 )-〔x^2 (y +z)+ y^2 (x +z) +z^2 (x +y)〕
=(x+y)(x-y)^2+(x+z)(x-z)^2+(y+z)(y-z)^2
>=0,
从而
2(x^3 +y^3 +z^3 )≥〔x^2 (y +z) y^2 (x +z) z^2 (x +y)〕.

追问

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Answer 2

2(x+y)²

Answer 3

是要化简么？

(x+y)2+(y+x)2
=(x+y)2+(x+y)2
=(x+y)4
=4x+4y