EDIT: rewritting the question to linear algebra to make it more accessible.
Denote by $\Delta([n])$ the set of all probability distributions over $\{1,2,\ldots,n\}$, that is: $$\Delta([n])=\{x\in[0,1]^n\mid \sum_{i=1}^n x_i=1\}$$
Let $A\in [0,1]^{n\times n}$ be a matrix, and let $x,y,z\in \Delta([n])$.
Does the following conditions:
- $\forall r\in\Delta([n]): r^tAy\leq x^tAy$
- $\forall r\in\Delta([n]): x^tA^tr\leq x^tA^ty$
- $\forall r\in\Delta([n]): r^tAz \leq \ \ z^tAz,\ \ z^tA^tr\ \ \leq \ \ z^tA^tz$
- $\forall i\in[n]: x_i+y_i > 0, z_i > 0$
Imply that $$x^tAy+x^tA^ty\geq 2\cdot z^tAz$$, or equivalently $$x^t(A+A^t)y\geq z^t(A+A^t)z$$?
For example, if
$A= \left( \begin{array}{ccc} 0.3 & 0.6 \\ 0.4 & 0.2 \\ \end{array} \right) $
Then $z=\left( \begin{array}{ccc} 0.8 \\ 0.2 \\ \end{array} \right)$ , $x=\left( \begin{array}{ccc} 1 \\ 0 \\ \end{array} \right)$ , $y=\left( \begin{array}{ccc} 0 \\ 1 \\ \end{array} \right)$
Satisfy the conditions and $$x^tAy+x^tA^ty = 0.6 + 0.4 > 0.36 \cdot 2 = 2\cdot z^tAz$$$$x^tAy+x^tA^ty = 0.6 + 0.4 > 0.36 + 0.36 = z^tAz\ + z^tA^tz$$
Notice that if condition (4) isn't true, then the claim doesn't hold, e.g.:
$A= \left( \begin{array}{ccc} 1 & 0 \\ 0 & 1 \\ \end{array} \right) $
And $z=\left( \begin{array}{ccc} 1 \\ 0 \\ \end{array} \right)$ , $x=y=\left( \begin{array}{ccc} 0.5 \\ 0.5 \\ \end{array} \right)$