Proof of Nonnegativity Inequality Prove the Inequality: 
$$\sum_{i,j}\left ( (PAQ)_{i,j}\frac{B_{i,j}^2}{A_{i,j}}-
(PBQ)_{i,j}B_{i,j}\right )  \geqslant 0$$
Given that:
$P$ and $Q$ are $n$x$n$ and $m$x$m$ symmetric matrices, 
$A$ and $B$ are $n$x$m$ matrices, and  
$A,B,P,Q$ are all non-negative (i.e. all elements $\geqslant 0$) and real.
In the above summation, $1 \leq i \leq n $ and $1 \leq j \leq m$. 
 A: Presumably $A$ is entrywise positive. Let $X$ be the entrywise positive square root of $A$, so that $A=X\circ X$. Let $B=X\circ Y$ for some entrywise nonnegative matrix $Y$. Then the sum in your inequality is the sum of all entries in the matrix
$$
(P(X \circ X)Q) \circ (Y \circ Y) - (P(X \circ Y)Q) (X \circ Y),
$$
which is equal to
\begin{align*}
&\mathrm{vec}(Y \circ Y)^T \mathrm{vec}(P(X \circ X)Q)
- \mathrm{vec}(X \circ Y)^T \mathrm{vec}(P(X \circ Y)Q)\\
=&\mathrm{vec}(Y \circ Y)^T (Q\otimes P) \mathrm{vec}(X \circ X)
- \mathrm{vec}(X \circ Y)^T (Q\otimes P) \mathrm{vec}(X \circ Y)\\
=&\mathrm{trace}\left\{\left[\mathrm{vec}(X \circ X) \mathrm{vec}(Y \circ Y)^T - \mathrm{vec}(X \circ Y) \mathrm{vec}(X \circ Y)^T\right] (Q\otimes P)\right\}\\
=&\mathrm{trace}(ZS),
\end{align*}
where $Z=\mathrm{vec}(X \circ X) \mathrm{vec}(Y \circ Y)^T - \mathrm{vec}(X \circ Y) \mathrm{vec}(X \circ Y)^T$ and $S=Q\otimes P$. Since $S$ is symmetric and nonnegative, if we define $x=\mathrm{vec}(X)$ and $y=\mathrm{vec}(Y)$, then
\begin{align*}
\mathrm{trace}(ZS)
&= \sum_{i<j} (z_{ij}+z_{ji})s_{ij} + \sum_i z_{ii}s_{ii}\\
&= \sum_{i<j} (x_i^2y_j^2-x_iy_ix_jy_j+x_j^2y_i^2-x_jy_jx_iy_i)s_{ij} + \sum_i (\underbrace{x_i^2y_i^2-x_iy_ix_iy_i}_{=0})s_{ii}\\
&= \sum_{i<j} (x_iy_j-x_jy_i)^2s_{ij} \ge0.
\end{align*}
