Question

两条线段的长度分别为 $a$ 个单位和 $b$ 个单位。则它们之间的正确关系为：

The lengths of two line segments are $a$ units and $b$ units respectively. Then the correct relation between them is:

$\dfrac{a+b}{2}\gt\sqrt{ab}$

$\dfrac{a+b}{2}\lt\sqrt{ab}$

$\dfrac{a+b}{2}=\sqrt{ab}$

$\dfrac{a+b}{2}\le\sqrt{ab}$

$\dfrac{a+b}{2}\ge\sqrt{ab}$

从非负的平方 $(a-b)^2$ 出发

Start with the nonnegative square $(a-b)^2$

当两条线段等长时，等号必须可以成立

Equality must remain possible when the two segment lengths are equal

因为 $(a-b)^2\ge0\text{，}$ 所以 $(a+b)^2\ge4ab$ 。两边都非负，因此 $\frac{a+b}{2}\ge\sqrt{ab}\text{。}$ 当 $a=b$ 时等号成立。

因此，正确答案是 E。

Since $(a-b)^2\ge0,$ we have $(a+b)^2\ge4ab.$ Both sides are nonnegative, so $\frac{a+b}{2}\ge\sqrt{ab}.$ Equality occurs when $a=b.$

Thus, the correct answer is E.