Saudi Arabia Mathematical Competitions

The inequality is equivalent to $2a^3+2b^3+2c^3+2abc\ge a^2(b+c)+b^2(c+a)+c^2(a+b)+2abc$, hence $$2a^3+2b^3+2c^3\ge ab(a+b)+bc(b+c)+ca(c+a)\text{\hspace{1em}(1)}$$ For any positive real numbers $x$ and $y$ we have $$x^3+y^3\ge xy(x+y)$$ Indeed, this inequality is equivalent to $x^2-xy+y^2\ge xy$, that is $(x-y{)}^2\ge 0$. We have equality in this inequality if and only if $x=y$. Applying the inequality above we get $a^3+b^3\ge ab(a+b)$, $b^3+c^3\ge bc(b+c)$, $c^3+a^3\ge ca(c+a)$. Adding these inequalities we obtain (1). We have equality in our inequality if and only if $a=b=c$.