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I'm working through a proof and am at a step where I start with the inequality $x > y$, then subtract 1 from both sides to get $x - 1 > y - 1$. From here what I need to do is get it so that the inequality becomes $x^2 - x > y^2 - y$. What can I do here? A friend of mine suggested multiplying the two inequalities together (referring to $x > y$ and $x - 1 > y - 1$), but I have never seen that done before or even heard of such a move. Is this possible?

jonc2006
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    Welcome to Mathematics Stack Exchange. When multiplying inequalities, be careful of the sign; multiplying by negative quantities reverses inequalities – J. W. Tanner Feb 08 '21 at 00:34
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    Regardless of whether $r,s$ positive or negative, if $t > 0$ and $r > s$, then $rt > st$. If $t < 0$ and $r > s$, then $rt < st$. The basis of all this is consideration that $r > s \iff (r-s) > 0$ and that if $p,q$ each positive, then so is $p \times q$. – user2661923 Feb 08 '21 at 00:38
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    Note that you are essentially claiming the function $f(x) = x^2 - x$ is always increasing. This is false for $x < 1/2$ (to the left of the parabola's vertex), so you would need more restrictions on your $x,y$ in order to have any hope of your inequality holding. – Brian Moehring Feb 08 '21 at 00:48

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