For each $d$, I have a matrix $M$ with values $$ M_{ij} = \begin{cases} \frac{4ij}{d} - \binom{2d}{d} & i \neq j & \\\\ \frac{4i^2}{d} - \binom{2d}{d} - \frac{\binom{2d}{d}}{\binom{d}{i}^{2}} & i = j \end{cases} $$
I want to show that, for every $d=2,3,\ldots$, the matrix is negative-definite.
An elegant answer has been provided by fedja, without needing to look at the determinants
My approach is to compute the determinant of the upper left-square matrix of size $k$, for each $k=1,2,\ldots,d$. The value for this is $$ D_k = (-1)^k\left[\frac{\binom{2d}{d}^k}{\prod_{i=1}^{k}\binom{d}{i}^2}\right] \left[\sum_{j=1}^{k}\left(\binom{d}{j}^2 - \frac{4j^2\binom{d}{j}^2}{d\binom{2d}{d}}\right) - \frac{4}{d\binom{2d}{d}} \sum_{1 \leq i < j \leq k}(j-i)^2\binom{d}{i}^2\binom{d}{j}^2\right] $$ Hence, $D_k$ has sign $(-1)^k$ if this expression is positive $$ \sum_{j=1}^{k}\left(\binom{d}{j}^2 - \frac{4j^2\binom{d}{j}^2}{d\binom{2d}{d}}\right) - \frac{4}{d\binom{2d}{d}} \sum_{1 \leq i < j \leq k}(j-i)^2\binom{d}{i}^2\binom{d}{j}^2 $$
I can re-write this as $$ \sum_{j=0}^{k}\binom{d}{j}^2 - \frac{2}{d\binom{2d}{d}} \sum_{0 \leq i,j \leq k}(j-i)^2\binom{d}{i}^2\binom{d}{j}^2\,, $$ and, then, after observing that $\sum_{i=0}^{d}i\binom{d}{i}^2 = \frac{d}{2}\binom{2d}{d}$, I can deduce that this will certainly be positive when the following sum is positive $$\sum_{0 \leq i, j \leq k}(i-(j-i)^2){\binom{d}{i}}^2{\binom{d}{j}}^2$$
Any advice/techniques would be appreciated.
