- 数学题证明:对于非负实数x、y,有(x^2+y^2)^2+x^3
- 证明:对于非负实数x、y,有(x^2+y^2)^2+x^3+y^3+4xy^2+x^2+y^2≥3x^3y+5xy^2+xy.
- (x^2+y^2)^2+x^3+y^3+4xy^2+x^2+y^2-(3x^3y+5xy^2+xy)
=x^4-3x^3y+2x^2y^2+y^4+x^3-xy^2+y^3+x^2-xy+y^2,
x^2-xy+y^2=(x-y/2)^2+3y^2/4>=0,①
f(x)=x^4-3x^3+2x^2+1(x>=0),
f'(x)=3x^3-9x^2+4x=0,
x1=0,x2,3=(9土√33)/6,
f(0)=1,f(x2)≈5,f(x3)≈1,
∴f(x)>0,
∴y^4*f(x/y)=x^4-3x^3y+2x^2y^2+y^4>=0.②
研究g(x)=x^3-x+1(x>=0),
g'(x)=3x^2-1=0,
x4=1/√3,
g(x4)=(9-2√3)/9>0,
g(0)=1,
∴g(x)>0,
∴y^3*g(x/y)=x^3-xy^2+y^3>=0.③
①+②+③,x^4-3x^3y+2x^2y^2+y^4+x^3-xy^2+y^3+x^2-xy+y^2>=0,
∴原式成立。