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?
 A: 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:

A: Your new hypothesis is still incorrect. Counterexample: for $A={\rm diag}(8,1)$ and $B=\begin{pmatrix}1&2\\2&8\end{pmatrix}$, we have $M=\frac{1}{2}(AB+BA)−2((AB)^{−1}+(BA)^{−1})^{−1} = \frac{49}{17}\begin{pmatrix}8&9\\9&8\end{pmatrix}$, whose two eigenvalues are $\frac{-49}{17}$ and $49$ resp., so $M$ is neither positive semidefinite nor negative semidefinite.
Next time, you may do some computational experiments first to see whether your hypothesis is violated. Counterexamples can usually be easily generated. The above counterexample of mine, for instance, is modified from some random output of the following Matlab code:

A=diag(rand(1,2));
D=diag(rand(1,2)); W=rand(2,2); Q=expm(W-W'); B=Q*D*Q';
M=0.5*(A*B+B*A)-2*inv(inv(A*B)+inv(B*A)); eig(M)

A: A slightly more rigorous argument is as follows: Note that the expressions can be rewritten as
$$
\begin{align}
0 &\stackrel{?}{\preceq} \frac{1}{2}(A+A^T) -2((A^{T})^{-1}+A^{-1})^{-1} \tag{Q}\\
&=\frac{1}{2}(A+A^T) -2 A^T(A+A^T)^{-1}A \tag{1}\\
&= \frac{1}{2}(A+A^T) -2 A(A+A^T)^{-1}A^T\tag{1'}\\
&= \frac{1}{2}(A+A^T)(A+A^T)^{-1}(A+A^T)-2 A^T(A+A^T)^{-1}A\\
&= \frac{1}{2}(A-A^T)(A+A^T)^{-1}(A-A^T)\\
&= -\frac{1}{2}(A-A^T)^T(A+A^T)^{-1}(A-A^T) \tag{2}
\end{align}
$$
Here (1) and (1') are equal and depending which side you are factoring the $A^{-1}$. And then we just recognize a "completing the square" step.
Here, note that (2) is nothing but a congruence transformation i.e. $P^TSP$ with skew symmetric $P=-P^T$, hence the sign of $S$ is preserved. Hence, whatever can be said about the (in)definiteness of $-(A+A^T)$, same holds for $(Q)$. Therefore, the answer is negative, (Q) does not hold in general. Replacing $A$ with $AB$ does not require any modification. 
One important detail is that even $A,B\succ 0$, this does not imply that $AB+BA\succ 0$ since $AB$ is not necessarily symmetric. (user1551 already constructed one example.)
