Question

下列说法中，唯一不正确的是：

Of the following statements, the only one that is incorrect is:

不等式两边同时增加、减少、乘以或除以（除数不为零）同一个正数后，不等式仍然成立。

An inequality will remain true after each side is increased, decreased, multiplied, or divided (zero excluded) by the same positive quantity.

两个不相等正数的算术平均数大于其几何平均数。

The arithmetic mean of two unequal positive quantities is greater than their geometric mean.

两个正数的和一定时，二者相等时乘积最大。

If the sum of two positive quantities is given, their product is largest when they are equal.

若 $a$ 和 $b$ 是不相等的正数，则 $\dfrac12(a^2+b^2)$ 大于 $\left[\dfrac12(a+b)\right]^2$ 。

If $a$ and $b$ are positive and unequal, $\dfrac12(a^2+b^2)$ is greater than $\left[\dfrac12(a+b)\right]^2.$

两个正数的乘积一定时，二者相等时和最大。

If the product of two positive quantities is given, their sum is greatest when they are equal.

答案：E

知识点：算术-几何平均不等式最优化

难度评级：1450

小提示：

在正乘积固定时，比较两因数相等时的和与两因数越来越不相等时的和

For a fixed positive product, compare the sum at equality with sums from increasingly unequal factors

大提示：

乘积固定时，算术平均值与几何平均值不等式给出和的最小值，而不是最大值

AM-GM gives a minimum, not a maximum, for the sum when the product is fixed

解答：

若 $uv=P\gt0$ ，则 $u+v\ge2\sqrt P$ ，并在 $u=v$ 时取等号。因此，两数相等时得到的是最小的和。令 $u$ 任意增大，同时令 $v=\frac{P}{u}$ 变小，二者之和可以无限增大，所以两数相等时和并非最大。

因此，正确答案是 E。

If $uv=P\gt0,$ then $u+v\ge2\sqrt P,$ with equality at $u=v.$ Thus equality gives the least possible sum. The sum can grow without bound by taking $u$ large and $v=\frac{P}{u}$ small, so it is not greatest at equality.

Thus, the correct answer is E.

1951 AMC 12 第 43 题

答案：E

难度评级：1450

小提示：

大提示：

解答：

其他年份的第 43 题