Fix $k\in \mathbb{N}$. Suppose $y_1,\ldots, y_{2k}$ are real numbers. Then $$\left(\sum_{1\le i_1<i_2<\cdots<i_k\le 2k} y_{i_1}y_{i_2}\cdots y_{i_k}\right)^2 \ge \binom{2k}{k}^2y_1y_2\cdots y_{2k}.$$
Note that the inequality follows easily by AM-GM when all $y_i$'s are non-negative reals. So, dealing with reals is the main challenge.
My idea: I was trying to use Newton's inequality. Let us write $d_k$ for the average of $k$-th elementary symmetric sum. So, the above inequality is just $d_{k}^2 \ge d_{2k}$. Indeed, for $k=1$, the above is precisely Newton's inequality. For $k\ge 2$, my initial idea was to use Newton repeatedly somehow. For example, I thought since $d_{2}^2 \ge d_1d_3$, squaring both sides and applying Newton again we get $$d_2^4 \ge d_1^2d_3^2 \ge d_0d_2d_2d_4 =d_2^2d_4$$ which boils down to $d_2^2 \ge d_4$. However, there is one mistake here. Since the variables are real here, apriori it is not clear if $d_2^2 \ge d_1d_3$ implies $d_2^4 \ge d_1^2d_3^2$. Since most of the other symmetric polynomial inequalities in the literature require non-negativity, I am not quite sure how to utilize them.
Any help/suggestions?
Regarding the source of the problem, this is something that comes up in my research. There i a potential chance that it is false, though from numerical computations it seems like it is true.