Posting the official solution here Let $F(x)=f_1(x)f_2(x)f_3(x)...f_n(x)$. We need to prove the concavity of $(F(x))^{\frac{1}{n}}$. By the assumption $F(\theta x +(1-\theta)y)\ge\prod_{i=1}^{n}[\theta f_i(x)+(1-\theta)f(y)]=\sum_{k=0}^{n}\theta^k(1-\theta)^{n-k}\sum f_{i_{1}}(x).....f_{i_{n}}(y)$ Where the second sum goes through all $\binom{n}{k}$ k subsets $(i_1,...,i_k)$ of $(1,2,.....,n)$ Now using Am Gm $\sum f_{i_{1}}(x)f_{i_2}(x).....f_{i_k}(x)\ge \binom{n}{k}F(x)^{\frac{k}{n}}F(y)^{\frac{n-k}{n}}$

We want to prove that the inequality $$t \sqrt[n]{\prod\limits_{k=1}^nf_k(x)}+(1-t)\sqrt[n]{\prod\limits_{k=1}^nf_k(y)} \le \sqrt[n]{\prod\limits_{k=1}^nf_k \left(tx+(1-t)y \right)}$$is true for all $x,y \in I$ and $t \in [0,1].$ This inequality is obviously true for $t \in \{0,1\}$ so we may assume that $t \in (0,1).$ Now let's fix $x,y$ and $t.$ Using the concavity of the functions $f_k,$ we get $$ \sqrt[n]{\prod\limits_{k=1}^n \left(tf_k(x)+(1-t)f_k(y) \right)}\le RHS,$$so it's enough to prove that $$t \sqrt[n]{\prod\limits_{k=1}^nf_k(x)}+(1-t)\sqrt[n]{\prod\limits_{k=1}^nf_k(y)} \le \sqrt[n]{\prod\limits_{k=1}^n \left(tf_k(x)+(1-t)f_k(y) \right)}$$Let's denote $u_k=tf_k(x)$ and $v_k=(1-t)f_k(y),~k= \overline{1,n}.$ The inequality becomes $$\sqrt[n]{ \prod\limits_{k=1}^n u_k}+\sqrt[n]{ \prod\limits_{k=1}^n v_k} \le \sqrt[n]{ \prod\limits_{k=1}^n (u_k+v_k)}.$$If there is some $k$ such that $u_k+v_k=0,$ then $u_k=v_k=0$ and the inequality is true. Otherwise, the inequality is equivalent to $$\sqrt[n]{\prod\limits_{k=1}^n \frac{u_k}{u_k+v_k}}+\sqrt[n]{\prod\limits_{k=1}^n \frac{v_k}{u_k+v_k}} \le 1.$$But from AM-GM we have $$\sqrt[n]{\prod\limits_{k=1}^n \frac{u_k}{u_k+v_k}}+\sqrt[n]{\prod\limits_{k=1}^n \frac{v_k}{u_k+v_k}} \le \frac{1}{n}\sum\limits_{k=1}^n \frac{u_k}{u_k+v_k}+\frac{1}{n}\sum\limits_{k=1}^n \frac{v_k}{u_k+v_k}=1,$$so we are done.