Thai Mathematical Olympiad

Let $a_6=3$ and $r_1,r_2,\dots ,r_6<0$ be all roots of $f(x)$. For each $k=1,2,\dots ,6$, we have $$a_k=(-1{)}^k\sum _{1\le j_1<\cdots <j_k\le 6}{r}_{j_1}{r}_{j_2}\cdots r_{j_k}=\sum _{1\le j_1<\cdots <j_k\le 6}|r_{j_1}||r_{j_2}|\cdots |r_{j_k}|$$ Using the AM-GM inequality, we have $$\begin{array}{rl}{a}_k& \ge \left(\genfrac{}{}{0ex}{}{6}{k}\right){\left[\prod _{1\le j_1<\cdots <j_k\le 6}|r_{j_1}||r_{j_2}|\cdots |r_{j_k}|\right]}^{1/\left(\genfrac{}{}{0ex}{}{6}{k}\right)}\\ & =\left(\genfrac{}{}{0ex}{}{6}{k}\right)(r_1{r}_2\cdots r_6{)}^{k/6}\\ & =3^{k/6}\left(\genfrac{}{}{0ex}{}{6}{k}\right)\\ & \ge \left(\genfrac{}{}{0ex}{}{6}{k}\right).\end{array}$$ Hence, $$\begin{array}{rl}f(2)& =2^6+2^5{a}_1+2^4{a}_2+2^3{a}_3+2^2{a}_4+2a_5+a_6\\ & \ge 2^6+2^5\left(\genfrac{}{}{0ex}{}{6}{1}\right)+2^4\left(\genfrac{}{}{0ex}{}{6}{2}\right)+2^3\left(\genfrac{}{}{0ex}{}{6}{3}\right)+2^2\left(\genfrac{}{}{0ex}{}{6}{4}\right)+2\left(\genfrac{}{}{0ex}{}{6}{5}\right)+\left(\genfrac{}{}{0ex}{}{6}{6}\right)\\ & =3^6\\ & =27^2.\end{array}$$