By the spectral theorem, if $A$ is symmetric it can be written as $Q^T\Lambda Q$ with $\Lambda $ the diagonal matrix whose entries are the eigenvalues of $A$ and $Q$ an orthogonal matrix. Then we have
$$v^TA v=v^TQ^T\Lambda Qv=w^T\Lambda w,$$
where $w=Qv$. Since $Q$ is orthogonal, we have
$$\|w\|=\langle w,w\rangle =\langle Qv,Qv\rangle =\langle v,Q^TQv\rangle =\langle v,v\rangle = \|v\|,$$
and thus
$$w^T\Lambda w=\sum_{i=1}^n\lambda_iw_i^2\leq \left(\sum_{i=1}^n\lambda_i\right)\|w\|^2=\operatorname{tr}(A)\|w\|^2=\operatorname{tr}(A)\|v\|^2,$$
because all the eigenvalues are non-negative, and $w_i^2\leq \sum_i w_i^2=\|w\|^2$.
Yes, it is positive semidefinite. This can be easily proved by mathematical induction on $n$.
To stress its size as well as its dependence on the parameters $b_i$s, let us denote the matrix generated by $(b_1,\ldots,b_n)\in[0,1]^n$ as described in your question as $A(b_1,\ldots,b_n)$. In the inductive step, let $b_i=\min(b_1,\ldots,b_n)$. Then
\begin{aligned}
A(b_1,\ldots,b_n)
&=\left[\begin{array}{c|c|c}
A(b_1,\,\ldots,\,b_{i-1})
&\begin{matrix}b_i\\ \vdots\\ b_i\end{matrix}
&\begin{matrix}b_i&\cdots&b_i\\ \vdots&&\vdots\\ b_i&\cdots&b_i\end{matrix}\\
\hline
\begin{matrix}b_i&\cdots&b_i\end{matrix}&b_i&\begin{matrix}b_i&\cdots&b_i\end{matrix}\\
\hline
\begin{matrix}b_i&\cdots&b_i\\ \vdots&&\vdots\\ b_i&\cdots&b_i\end{matrix}
&\begin{matrix}b_i\\ \vdots\\ b_i\end{matrix}
&A(b_{i+1},\,\ldots,\,b_n)
\end{array}\right]\\
&=\left[\begin{array}{c|c|c}
A(b_1-b_i,\,\ldots,b_{i-1}-b_i)&0&0\\
\hline
0&0&0\\
\hline
0&0&A(b_{i+1}-b_i,\,\ldots,\,b_n-b_i)
\end{array}\right]+b_iE
\end{aligned}
where $E$ denotes the all-one matrix. Since $b_k-b_i\in[0,1]$ for every $k$, we see that $A(b_1,\ldots,b_n)$ is positive semidefinite, by induction assumption.
Best Answer
Suppose that $A = [a_{ij}]_{i,j=1}^n$ is such that $a_{ii} < 0$ for some $i$. Let $e_i$ be the $i$th standard basis vector; that is, $$ e_i = (\overbrace{0,\cdots,0}^{i-1},1,0,\dots,0) $$ then $e_i^T Ae_i = a_{ii} < 0$, which means that $A$ is not positive semidefinite.
So, if $A$ is positive semidefinite, then all diagonal elements are non-negative, which means that the trace is non-negative.