Markov's Inequality Proof confusion I'm going through the proof of Markov's Inequality, defined as

For a non-negative random variable $X$ with expectation $E(X)=\mu$, and any $\alpha > 0, P[X \geq \alpha] \leq \frac{E(X)}{\alpha}$

So, to understand what this was trying to say in the first place, I rephrased it as "the probability that non-negative r.v. $X$ takes on a value greater than $\alpha$ is upper bounded by $\frac{E(X)}{\alpha}$".
Now, my book goes into the proof in the following manner:
$E(X) = \sum_a a \times P[X=a]$ this line makes perfect sense
$E(X) \geq \sum_{a \geq \alpha} a \times P[X=a]$ why did they introduce an inequality here? The book also says that this is a crucial step where we have used the fact that $X$ takes on only non-negative values. I don't understand where that property is being used. I tried plugging in an r.v. that took negative values and didn't see anything out of the ordinary - why is it a constraint in this theorem?
$E(X) \geq \alpha \sum_{a \geq \alpha} P[X=a]$ how can you pull out $\alpha$ out of the summation?
$E(X) = \alpha P[X \geq \alpha]$ how did the inequality change places?

We have $E(X)=\sum_a aP(X=a)=\sum_{a:a<\alpha}aP(X=a)+\sum_{a:a\geq \alpha}aP(X=a)$
Since $X\geq0$ we have the values that $X$ takes are non-negatie, so $aP(X=a)\geq0$ for all $a$ (If $a$ is negative then $P(X=a)=0$ so $aP(X=a)=0$). Therefore, $\sum_{a:a<\alpha}aP(X=a)\geq0$
Thus $E(X)\geq \sum_{a:a\geq\alpha}aP(X=a)$
In this final summation, each $a\geq\alpha$ so $aP(X=a)\geq \alpha P(X=a)$ thus $\sum_{a:a\geq\alpha}aP(X=a)\geq \alpha\sum_{a:a\geq \alpha}P(X=a)=\alpha P(X\geq\alpha)$
Hence you finally get $E(X)\geq \alpha P(X\geq \alpha)$ for any $\alpha>0$.
