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**:

 1. $\forall r\in\Delta([n]): r^tAy\leq x^tAy$
 2. $\forall r\in\Delta([n]): x^tA^tr\leq x^tA^ty$
 3. $\forall r\in\Delta([n]): r^tAz,\ \ z^tA^tr\ \  \leq \ \ z^tAz$
 4. $\forall i\in[n]: x_i+y_i > 0, z_i > 0$

> **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 $$x^tAy+x^tA^ty = 0.6 + 0.4 > 0.36 \cdot 2 = 2\cdot z^tAz$$

---

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 $z=\left( \begin{array}{ccc}
1 \\
0  \\
\end{array} \right)$ , $x=y=\left( \begin{array}{ccc}
0.5 \\
0.5  \\
\end{array} \right)$