Let $$\mathbf{E} = \left[ \begin{array}{ccc} 1 & a_1 & b_1 \\ 1 & a_2 & b_2 \\ \vdots & \vdots & \vdots \\ 1 & a_n & b_n \end{array} \right] $$ where $0 < a_i,b_i < 1$, $b_i < a_i$ for $i=1,...n$, and $a_{i+1} < a_{i}$ for $i=1,...,n-1$.
Is there a proof that, for $n>2$, $\mathbf{A} := \mathbf{E}'\mathbf{E}$ is positive (semi-) definite (I've done endless simulations and the suggestion is that it is, plus, in the context of the application it makes sense that it is)? Reference to book or paper would be ideal.
Here is what I have so far:
Since $$ \mathbf{A} = \left[ \begin{array}{ccc} N & \sum a_i & \sum b_i \\ \sum a_i & \sum a_i^2 & \sum a_i b_i \\ \sum b_i & \sum a_i b_i & \sum b_i^2 \end{array} \right] $$ where $ \sum x_i := \sum_{i=1}^n x_i$, clearly $\mathbf{A}$ is symmetric. Further, as regards the principal minors, we have, first, $N>0$ and second (by Chebyshev sum inequality), $N \sum a_i^2 - \left(\sum a_i \right)^2 \geq 0$, so it remains to show $|\mathbf{A}| \geq 0$ (note that since the $b_i$ are not ordered Chebyshev is not available for the $b_i$).
Probably unsurprisingly, whether I use Schur complement, Descartes change-of-sign rule on the characteristic equation, or the determinant itself, the proof seems to lead to the same requirement, being to show that $$ N\sum a_i^2 \sum b_i^2 + 2 \sum a_i \sum b_i \sum a_i b_i - \left[ N\left( \sum a_i b_i \right)^2 + \left( \sum a_i \right)^2\sum b_i^2 + \sum a_i^2 \left( \sum b_i \right) ^2 \right] \geq 0. $$
It fails if the columns of $E$ are linearly dependent. This would include all $a_i = b_i,$ for example. This would also include all $b_i = (1/2) a_i.$ In general, the existence of a triple of real constants $\gamma, \alpha, \beta$ such that $\gamma + \alpha a_i + \beta b_i = 0$ for all $i$ would mean your $E^T E$ has rank two or one, is positive semidefinite but not definite. Indeed, take column vector $$ v = \left( \begin{array}{c} \gamma \\ \alpha \\ \beta \end{array} \right) $$
Then $$ v^T E^T E v = 0. $$