Inequality with $a\le b\le c$ and $a+b+c=1$

Question

Let $a,b,c$ be positive reals with $a\le b\le c$ and $a+b+c=1$. Is it true that $$ 3a(b+c)+2bc \le 3(a+b)(b+c)(c+a)\,\,? $$

Answer 1

$$3(a+b)(a+c)(b+c)-(3ab+3ac+2bc)(a+b+c)=(b+c-2a)bc\geq0.$$

Answer 2

we get $$3(a+b)(b+c)(c+a)-3a(b+c)-2bc=3\,{a}^{2}b+3\,{a}^{2}c+3\,a{b}^{2}+6\,abc+3\,a{c}^{2}+3\,{b}^{2}c+3\, b{c}^{2}-3\,ab-3\,ac-2\,bc $$ abd with $$c=1-a-b$$ we obtain $$b(3a-1)(a+b-1)$$ Can you proceed?