How is inequality condition reduced to this? From Convex Optimization:
How is the condition in the red box reduced to this?


 A: After having derived the condition $\nabla f_0(x) \succeq 0$, you know $\nabla f_0(x)^T y \geq 0$. Since the inequality has to hold for all $y \succeq 0$, it also has to hold for the $y$ which attains the minimum of $\nabla f_0(x)^T y $: $y=0$. For $y=0$ you get $-\nabla f_0(x)^Tx \geq 0$.
So you need all three conditions: $x \succeq 0$, $\nabla f_0(x) \succeq 0$, and $-\nabla f_0(x)^Tx \geq 0$.
A: Think about the optimality condition. You have a $ y \succeq 0$ hypothesis. But you need to check 
$$ \nabla f_0(x)^T(y-x) \geq 0$$
So as $\succeq$ is understood entry-wise, this means every $y_i \geq 0$ and so $y_i - x_i \geq -x_i$, so
$$ \nabla f_0(x)^T(y-x) \geq \nabla f_0(x)^T(-x)  \: \: \forall y \succeq 0$$ 
$y\succeq 0$ includes the case $ y = 0$ for which the optimality condition means $\nabla f_0(x)^T(-x) \geq 0 $
Do you see why it is sufficient? 
$$ \nabla f_0(x)^T(y-x) \geq \nabla f_0(x)^T(-x) \geq 0 $$ 
A: I figured it out.  Since it's for all $y$, it doesn't matter what $y$ we choose so it just matters that $-()$ is positive because it's a subtraction.
