For what A, If $A+A^T>0$ then $A^2+A^{2T}>0$? let me know if I am wrong with the next with a real square matrix A.
$A+A^T = \sqrt{A^2+A^{2T}+AA^T+A^TA} > 0$
This square root exists right? And because of this, the sum of all its elements are also a positive definite matrix.
So, because $AA^T > 0$, then $-(A^2+A^{2T}) < AA^T+A^TA$, so the left hand could be negative.
Is there any way for checking the expression of the question only looking at A without computing the squares?
UPDATE
With > 0 I mean a positive definite matrix.
 A: If by
$$A+A^T = \sqrt{A^2+A^{2T}+AA^T+A^TA}$$
you mean that the square of $A+A^T$ is $A^2+A^{2T}+AA^T+A^TA$ then yes, that square root does exist.
As for the question in the title we never compare matrices with numbers so $A + A^T > 0$ doesn't mean anything.  From your question I get the impression that you mean to say that the entries of the matrix $A + A^T$ are all positive numbers.  In that case it's not in general true that $A^2 + A^{2T}$ has positive entries when $A + A^T$ does.  For example take
$$A = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}$$
Then you'll find that
$$A + A^T = \begin{bmatrix} 2 & 1 \\ 1 & 6 \end{bmatrix}$$
has all positive entries but
$$A^2 + A^{2T} = \begin{bmatrix} -2 & 8 \\ 8 & 14 \end{bmatrix}$$
does not.  I suspect you'll just need to compute the squares to know if any particular $A$ gives a matrix $A^2 + A^{2T}$ with all positive entries.
Edit: If you mean does $A + A^T$ positive definite imply $A^2 + A^{2T}$ is then that is false as well.  Take
$$A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$
Then $A + A^T$ has eigenvalues $1$ and $3$ so it is positive definite, but $A^2 + A^{2T}$ has $0$ as an eigenvalue, so it is not.
