Problem inequality

Question

Let $a,b,c>0$ and $a+b+c=3$. Prove that $$a^3+b^3+c^3+2abc \ge a^2+b^2+c^2+2$$ I can solve it, I let $(a+b+c;ab+bc+ca;abcc)=(p;q;r)$. I need another solution

Answer 1

Can't edit my comment & links don't post correctly -- so, posting it here again as an answer with minor typo corrected. Consider $a^3+b^3+c^3+2abc−a^2−b^2−c^2−2$. You can use Lagrange multipliers to find the minima. It is 0, at a=1, b=1, c=1.

Try Minimize[{x^3+y^3+z^3+2xyz-x^2-y^2-z^2-2, x+y+z=3, x>0,y>0,z>0},{x,y,z}] at Mathematica website.

Answer 2

We need to prove that $$27(a^3+b^3+c^3+2abc)\geq9(a+b+c)(a^2+b^2+c^2)+2(a+b+c)^3$$ or $$\sum_{cyc}(27a^3+18abc-9a^3-9a^2b-9a^2c-2a^3-6a^2b-6a^2c-4abc)\geq0$$ or $$\sum_{cyc}(16a^3-15a^2b-15a^2c+14abc)\geq0$$ or $$15\sum_{cyc}(a^3-a^2b-a^2c+abc)+\sum_{cyc}(a^3-abc)\geq0,$$ which is true by Schur and AM-GM.