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Let matrix $X$ satisfy a differential equation $ \dot X = f(t,X) $ where right side is real and symmetric. Let $X(0) = M = M^{T} \succeq 0$ and matrix $Y$ satisfy differential inequality $ \dot Y \succeq f(t,Y), \;\;\; Y(0) = M $ where $A \succeq B$ means that for any vector $v$ we have $v^{T}Av \geq v^{T}Bv$. Is it true that $ Y(t) \succeq X(t) $ for $t> 0$? Function $f$ is sufficiently smooth.

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    @ErickWong thank you for information about notation, I didn't see it.2012-06-13

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I think this is a counterexample, even with a time-independent equation in which the matrices are diagonal. Let $f((a,b))=((0,-2a))$ where $((a,b))$ means the 2 by 2 diagonal matrix with diagonal entries $a,b$. Starting with $M=((0,0))$, we have $X=((0,0))$ for all times. On the other hand, $Y(t)=((t,-t^2))$ satisfies $\dot{Y}(t)=((1,-2t))\ge ((0,-2t))=f(Y)$.