Question

非零实数 $x$ 、 $y$ 、 $a$ 和 $b$ 满足 $x < a$ 且 $y < b$ 。下面四个不等式中有多少个必然成立？

(I) $x + y \lt a + b$

(II) $x - y \lt a - b$

(III) $xy \lt ab$

(IV) $\dfrac{x}{y} \lt \dfrac{a}{b}$

Nonzero real numbers $x,$ $y,$ $a,$ and $b$ satisfy $x < a$ and $y < b.$ How many of the following inequalities must be true?

(I) $x + y \lt a + b$

(II) $x - y \lt a - b$

(III) $xy \lt ab$

(IV) $\dfrac{x}{y} \lt \dfrac{a}{b}$

$0$

$1$

$2$

$3$

$4$

把两个不等式相加，得到

所以 (I) 必然成立。 $x + y \lt a + b,$

不等式不能直接相减，所以 (II) 不一定成立。

例如取 $x = 1$ $y = 1$ $a = 2$ $b = 3$ ，就会得到错误的不等式 $0 \lt -1$ 。

(III) 也不一定成立，因为 $x$ 和 $y$ 可能都是负数。

取 $x = -3$ $y = -2$ $a = 1$ $b = 1$ 则 $xy = 6$ ，而 $ab = 1$ ，所以 (III) 不成立。

(IV) 也有同样的问题。使用同一组数，有 $\dfrac{x}{y} = 1.5$ ，而 $\dfrac{a}{b} = 1$ 。

因此只有 (I) 必然成立。

所以正确答案是 B。

Adding the two inequalities together gets us

$x + y \lt a + b,$ which shows that (I) is correct.

One cannot subtract inequalities, which means that (II) is not necessarily true.

Consider $x = 1,$ $y = 1,$ $a = 2,$ and $b = 3$ as a counter-example. This would give us $0 \lt -1.$

(III) is also not always true, since $x$ and $y$ might be negative numbers.

Let $x = -3,$ $y = -2,$ $a = 1,$ and $b = 1.$ Then $xy = 6$ and $ab = 1$ which shows that (III) is wrong.

The same thing occurs with (IV) . Using the same values as above, we have $\dfrac{x}{y} = 1.5$ and $\dfrac{a}{b} = 1.$

This shows that (I) is the only true statement.

Thus, B is the correct answer.