The inequality to prove rewrites as $ A = (\sum x_k^2y_k)(\sum y_k) - (\sum x_ky_k)^2\ge B = \frac14(\sum y_k)(\sum y_kx_kx_{k - 1})$. From the identity $ (\sum a_i^2)(\sum b_i^2) - (\sum a_ib_i)^2 = \displaystyle\sum_{i < j}(a_ib_j - a_jb_i)^2$, we have $ A = \sum_{i < j}y_iy_j(x_i - x_j)^2$. We are going to resort now to induction. For $ n = 1,2$ the inequality can be verified easily. For the step of the induction, it is sufficient to prove that $ y_n\displaystyle\sum_{k = 1}^{n - 1}y_k(x_n - x_k)^2\ge \frac {y_nx_nx_{n - 1}(y_1 + \ldots + y_{n - 1} + y_n)}4 +$ $ y_n\cdot\frac {y_2x_2x_1 + \ldots + y_{n - 1}x_{n - 1}x_{n - 2}}4$. Since $ x_k\le x_n$, it is enough to prove the last inequality for $ x_1 = \frac {x_2}2 = \ldots = \frac {x_n}n$. Actually we can assume $ x_i = i$, $ i = 1,2,\ldots,n$. Hence we are to prove $ \displaystyle\boxed{\sum_{k = 1}^{n - 1}y_k(n - k)^2\ge\frac {n(n - 1)(y_1 + \ldots + y_n)}4 + \frac {y_2(2^2 - 2) + \ldots + y_{n - 1}[(n - 1)^2 - (n - 1)]}4}$. Let ${ Y = y_1 + \ldots + y_{n - 1}}$. From $ y_n\le y_{n - 1}\le\ldots\le y_1$ and rearrangement inequality we deduce the following: A. $ y_1 + \ldots + y_n\le \frac n{n - 1}Y$. B. $ y_2 + \ldots + y_{n - 1}\le \frac {n - 2}{n - 1}Y$. C. $ \displaystyle\sum_{k = 1}^{n - 1}y_k(n - k)^2\ge \frac {Y[1^2 + 2^2 + \ldots + (n - 1)^2]}{n - 1} = \frac {n(2n - 1)}6Y$. D. $ E = y_2(2^2 - 2) + \ldots + y_{n - 1}[(n - 1)^2 - (n - 1)]\le$ $ \frac {(y_2 + \ldots + y_{n - 1})[1^2 - 1 + 2^2 - 2 + \ldots + (n - 1)^2 - (n - 1)]}{n - 2}$. E. $ E\le\frac Y{n - 1}[\frac {(n - 1)n(2n - 1) - 3n(n - 1)}6]$. Hence, it would be enough to have that $ \frac {n(2n - 1)}6Y = \frac {n^2}4Y + Y[\frac {n(2n - 1) - 3n}{24}]$, which is an identity.