Is $2A^2 \geq AB+BA$ when $A\geq B\geq 0$ ? Always true? Let $A$ and $B$ be real (symmetric) and positive definite. It follows that $AB+BA$ is not necessarily positive definite (it can be indefinite, negative definite or positive definite). 

But now suppose $A\geq B$. Can one always say $2A^2 \geq AB+BA$? 

Since this ordering implies $2A^2-AB-BA\geq0$ and adding indefinite/neg-def/pos-def to a pos-def matrix may still be pos-def I am supposing the question is well-defined.
No. Let me reuse an old counterexample, where $\epsilon&gt;0$ is small:
\begin{align*}
A&amp;=\pmatrix{2+\epsilon&amp;1\\ 1&amp;3},\\
B&amp;=\pmatrix{1\\ &amp;2},\\
2A^2-AB-BA&amp;=\pmatrix{2(1+\epsilon)(2+\epsilon)+2&amp;7+2\epsilon\\ 7+2\epsilon&amp;8}\approx\pmatrix{6&amp;7\\ 7&amp;8}\not\succeq0.
\end{align*}
Note that $(A-B)^2 \ge 0$ in the sense that you defined earlier. Now
$(A-B)^2 = A^2 - AB - BA + B^2$ and thus 
$$
A^2 + B^2 \ge AB + BA\, .
$$
Now use $A^2 \ge B^2$ to arrive at the conclusion.