You are asked to "Offer a proof of the following": Suppose that $X$ is a random variable, $X \in \mathbb{R}$. If $7X^3+5X \ge X^2+8$, then $X \ge 0$
Proof 1: $X(7X^2+5) \ge (X^2+8)$. $X \ne 0$ because $(0)(5) \ge 8$ is not true. $X$ cannot be less than zero because (negative)(positive) >= (positive) is also not true. Therefore, $X \gt 0$. Therefore, $X \ge 0$
Proof 2: $7X^3 - X^2 + 5X - 8 \ge 0$, solving for the real roots, we find $X \ge 0.8593196$. Therefore, $X\ge 0$
My question is on the last step of either proof. When asked to perform a proof, are you only saying that I have proven a subset, $X>0$ or $X \ge 0.8593196$, therefore the superset is true, that $X \ge 0$, OR are you making a statement about the domain of $X$, that zero is an allowed value of $X$ which makes the last conclusion false so you cannot prove $X\ge0$.
Is there an authoritative reference for this? Perhaps a manual on proofs?
Note: This is not a homework problem, I have tried to distill the question to help elucidate the point.
Edit
Perhaps a different way of asking the question is whether a proof like this is asking if there exists at least one $X$ that makes both statements true, or if we are to prove that the support of $X$ is on $[0,\infty)$?
There's an even more elegant way, without a single calculation - proving by contradiction:
Let $7X^3+5X\ge X^2+8$ and assume $X<0$. The term $X^2+8$ is bound to be positive. As $X$ is negative, $X^3$ is also negative and thus the term $7X^3+5X$ is promised to be negative. That is, if we assume $X<0$ we reach a contradiction because $7X^3+5X< X^2+8$. We can conclude that if $7X^3+5X\ge X^2+8$ then $X\ge0$. $\blacksquare$
Generally speaking, when we need to prove that A leads to B ("if A then B"), there are 3 major types of mathematical proofs:
There are, of course, many other types of proofs, but these are the main 3 types AFAMKG.