Whether this matrix is positive definite Let $A$ be a nonsingular real square matrix. Is it true that the matrix 
$$\frac{1}{2}(A+A')-2(A^{-1}+(A^{-1})')^{-1}$$ 
is positive semidefinite?
Here, $A'$ denotes the transpose of $A$.
Edited 
Let $A,B$ be positive definite matrices of the same size, is it true that $\frac{1}{2}(AB+BA)−2((AB)^{−1}+(BA)^{−1})^{−1}$ is positive semidefinite?

Added: This answer the updated question. 
Your updated conjecture is also false. Consider:
$$
  A = \left(
\begin{array}{cc}
 10 & -5 \\
 9 & 3 \\
\end{array}
\right) \qquad\qquad\qquad 
 B = \left(
\begin{array}{cc}
 1 & -6 \\
 8 & 8 \\
\end{array}
\right)
$$
Then
$$
  \frac{1}{2}(A.B + B.A) - 2 ( (A.B)^{-1} + (B.A)^{-1} )^{-1} =  \frac{1}{7474} \left(
\begin{array}{cc}
 -98938 & -164451 \\
 247345 & -61502 \\
\end{array}
\right)
$$
The matrix above is negative-definite.

Added: This portion answers the first variant of the problem:

No, it is not true. Counterexample:
$$
  A = \left(
\begin{array}{ccc}
 3 & 3 & -4 \\
 1 & -2 & 0 \\
 -1 & -2 & -3 \\
\end{array}
\right)
$$
Then 
$$
   \frac{1}{2}( A + A^t) - 2 ( A^{-1} + (A^{-1})^t)^{-1} = \left(
\begin{array}{ccc}
 \frac{127}{198} & \frac{4}{33} & -\frac{163}{198} \\
 \frac{4}{33} & -\frac{2}{11} & \frac{5}{33} \\
 -\frac{163}{198} & \frac{5}{33} & \frac{59}{99} \\
\end{array}
\right)
$$
It eigenvalues are, approximately,  $1.44$, $-0.387$ and zero. 
Added: In order to create such a counterexample I have used Mathematica:
f[a_] := 1/2 (a + Transpose[a]) - 
  2 Inverse[# + Transpose[#] &[Inverse[a]]]

Now this generates random integer-valued matrices until a non-degenerate one is generated with the combination in question having eigenvalues of opposite signs:
While[True,
  While[Det[a = RandomInteger[{-4, 4}, {2, 2}]] == 0, Null];
  If[ Intersection[Sign[Eigenvalues[f[a]]], {-1, 1}] == {-1, 1}, 
   Break[]]
  ];

Here is a screenshot:

