TITLE: Prove that if $x+y \geq 2$, where x and y are real numbers, then, x $\geq 1 \vee y \geq 1$.
QUESTION [0 upvotes]: I understand that one can prove this by using contraposition. The point that got me unsatisfied with this solution is that I have no idea about why one wouldn't use the Direct Proof Method when proving this? If possible, could you please use the direct proof method for the solution of this question and well, if impossible, explain the reason behind the uselesness of the direct proof method in this example?

REPLY [2 votes]: We have $x-1+y-1\geq0$.
Let $y-1<0$.
Thus, $x-1\geq-(y-1)>0$, which gives $x-1\geq0,$ which we need.
But, $y-1\geq0$ gives, which we need again.
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