Selection and Training Session

By the Cauchy-Buniakowski inequality, $$({\sqrt{a}}^2+{\sqrt{b}}^2+{\sqrt{c}}^2)((\sqrt{a}(b+c){)}^2+(\sqrt{b}(c+a){)}^2+(\sqrt{c}(a+b){)}^2)\ge (a(b+c)+b(c+a)+c(a+b){)}^2=4(ab+bc+ca{)}^2.$$ Hence, $$(a+b+c)(a(b+c{)}^2+b(c+a{)}^2+c(a+b{)}^2)\ge 4(ab+bc+ca{)}^2\ge 4(ab+bc+ca)(a+b+c)\iff (a(b+c{)}^2+b(c+a{)}^2+c(a+b{)}^2)\ge 4(ab+bc+ca).$$ It remains to note that $$a(b+c{)}^2+b(c+a{)}^2+c(a+b{)}^2=(a+b+c)(ab+bc+ca)+3abc,$$ which gives the required inequality.

Alternative solution:

Let ${\sigma }_1=a+b+c$, ${\sigma }_2=ab+bc+ca$, ${\sigma }_3=abc$. By the problem condition, ${\sigma }_2\ge {\sigma }_1$. Since ${\sigma }_1^2\ge 3{\sigma }_2$ (well-known inequality), we have ${\sigma }_1^2\ge 3{\sigma }_2\ge 3{\sigma }_1$, i.e. ${\sigma }_1\ge 3$. Now we use Schur's inequality ${\sigma }_1^3+9{\sigma }_3\ge 4{\sigma }_1{\sigma }_2$. We have ${\sigma }_1^2{\sigma }_2\ge {\sigma }_1^3$ and $3{\sigma }_1{\sigma }_3\ge 9{\sigma }_3$, hence ${\sigma }_1^2{\sigma }_2+3{\sigma }_1{\sigma }_3\ge 4{\sigma }_1{\sigma }_2$ and then ${\sigma }_1{\sigma }_2+3{\sigma }_3\ge 4{\sigma }_2$, as required.

Alternative solution:

Since $ab+bc+ca\ge a+b+c$, it suffices to prove the inequality $(a+b+c{)}^2(ab+bc+ca)+3abc(a+b+c)\ge 4(ab+bc+ca{)}^2$. The last inequality can be easily transformed to $$a^{3}b+b^{3}a+a^{3}c+c^{3}a+b^{3}c+c^{3}b\ge 2(a^2{b}^2+b^2{c}^2+c^2{a}^2),$$ which is true by Muirhead's theorem.