# Basic Question on Inequalities

Let x and y be real numbers

And I know that

$$x-y < x*y < 0$$

Then which one is true?

(I) $$x^2-(x*y)<0$$

(II) $$y(x-1)<0$$

(III) $$y+ \frac{y}{x}<1$$

My Solution is

(III) is wrong. Why?

If $$x-y < xy$$ then $$x < y+ xy$$ $$\frac{x}{x} < \frac{y}{x} + \frac{xy}{x}$$ $$1< \frac{y}{x} + y$$

(II) is true.

$$x-2y < x-y < xy -y =(x-1)y$$

I have no idea for (I).

However, the book’s answer key says that (II) and (III) are true. Why? I don’t see this. Please help me to prove that. Thanks a lot.

• You can only divide an inequality by a positive number. If you divide by a neagtive number the inequality gets reversed. Ex: $1<2$ but $\frac 1 {-1} >\frac 2 {-1}$. Jul 8 '20 at 10:10
• Thanks. Then (III) is true. Well, please show me the part (I). I have no idea for (I). @KaviRamaMurthy
– 1190
Jul 8 '20 at 10:14
• Since $xy \lt 0$, then one of $x$ and $y$ is negative and the other is positive. With $x - y \lt 0$, this determines which specific value is negative & which is positive. Jul 8 '20 at 10:14
• So, $x<y$. Then x is Negative, y is positive. Is it right? @JohnOmielan
– 1190
Jul 8 '20 at 10:16
• @B11b Yes, $x \lt 0$ and $y \gt 0$. Also, for (I), note $x^2 \gt 0$ and $xy \lt 0 \implies -xy \gt 0$, so what does that say about the sum of those $2$ terms? Jul 8 '20 at 10:18

(I) is wrong. $$x^2 -xy < 0$$ and $$xy < 0$$ implies that $$x^2 < xy<0$$ but $$x^2 \ge 0$$.
(II): If $$x>0$$, then $$y<0$$ since $$xy<0$$. However, this would imply that $$x-y >0$$ which is a contradiction. Hence we must have $$x<0$$, $$y>0$$. Since $$x-1 < 0$$ and $$y>0$$, we must have $$(x-1)y < 0$$.
(III): As argued earlier, $$x<0$$,
Hence $$1>\frac{y}{x}+y$$