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. $\forall r\in\Delta([n]): r^tAz \leq \ \ z^tAz,\ \ z^tA^tr\ \ \leq \ \ z^tA^tz$
4. $\forall i\in[n]: x\_i+y\_i > 0, z\_i > 0$



> 
> **Imply that $$x^t(A+A^t)y\geq z^t(A+A^t)z$$?**
> 
> 
> 




---


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 $$x^tAy+x^tA^ty = 0.6 + 0.4 > 0.36 + 0.36 = z^tAz\ + z^tA^tz$$




---


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



(linear algebra) - Can a symmetric equilibrium achive higher social-welfare than some equilibrium with the same support?
I am the aforementioned economist. 
I figured out a counterexample for the non-singular case:


Let 
$$
A = \left( \begin{array}{ccc}
0 & 7 & 0 \\
7 & 0 & 0 \\
3 & 3 & 1 \end{array} \right)
$$


$$
z = \left( \begin{array}{ccc}
\frac{1}{3}, & \frac{1}{3}, & \frac{1}{3} \end{array} \right)
$$


$$
x = \left( \begin{array}{ccc}
\frac 15, & 0, & \frac45 \end{array} \right)
$$


$$
y = \left( \begin{array}{ccc}
0, & \frac 15, & \frac45 \end{array} \right)
$$


The expected payoff is $2\cdot \frac 73$ in the symmetric equilibrium and $\frac 75 + \frac 75$ in the other one. As far as I can tell there is no 2x2 non-singular counterexample.


