Question

已知a+b+c=1求证 a3+b3+c3>=1/3(a2+b2+c2)

Answer 1

3（a^3+b^3+c^3)-(a^2+b^2+c^2)(a+b+c)

=3（a^3+b^3+c^3)-(a^3+b^3+c^3+a^2b+a^2c+ab^2+b^2c+ac^2+bc^2)

=2a^3+2b^3+2c^3-a^2b-a^2c-ab^2-b^2c-ac^2-bc^2

=(a^3+b^3-a^2b-ab^2)+(b^3+c^3-bc^2-b^2c)+(c^3+a^3-ca^2-ac^2)

=(a+b)(a^2-2ab+b^2)+(b+c)(b^2+c^2-2bc)+(a+c)(a^2+c^2-2ac)

=(a+b)(a-b)^2+(b+c)(b-c)^2+(a+c)(a-c)^2≥0

∴a^3+b^3+c^3≥1/3(a^2+b^2+c^2)(a+b+c)=1/3(a^2+b^2+c^2)