- 已知$x_0$是函数$f(x) = a + 2b\sqrt{x} + e^{\frac{x}{2}}$的一个零点，且$x_0 \in [1, e]$，则$a^2 + b^2$的最小值
解：思路一
依题可知$a + 2b\sqrt{x_0} + e^{\frac{x_0}{2}} = 0$
$\therefore a + 2b\sqrt{x_0} = -e^{\frac{x_0}{2}}$
$\because (a^2 + b^2)(1 + 4x) \geq (a + 2b\sqrt{x_0})^2$（当且仅当$b = 2a\sqrt{x}$时等号成立）
$\therefore a^2 + b^2 \geq \frac{\left( e^{\frac{x_0}{2}} \right)^2}{1 + 4x_0} = \frac{e^{x_0}}{1 + 4x_0}$
令$g(x) = \frac{e^x}{1 + 4x}$，则$g'(x) = \frac{e^x(4x - 3)}{(1 + 4x)^2}$
$\therefore x > \frac{3}{4}$时，$g(x)$单增
$\therefore x \in [1, e]$时，$g(x)_{\min} = g(1) = \frac{e}{5}$
$\therefore a^2 + b^2 \geq \frac{e}{5}$
思路二
$\because$依题可知$a + 2b\sqrt{x_0} + e^{\frac{x_0}{2}} = 0$
$\therefore$点$(a, b)$在直线$x + 2\sqrt{x_0}y + e^{\frac{x_0}{2}} = 0$上
$\sqrt{a^2 + b^2} = \sqrt{(a - 0)^2 + (b - 0)^2}$可以看成是点$(a, b)$到原点$(0, 0)$的距离
$\frac{\left| e^{\frac{x_0}{2}} \right|}{\sqrt{1 + 4x_0}}$可以看成是原点$(0, 0)$到直线$x + 2\sqrt{x_0}y + e^{\frac{x_0}{2}} = 0$的距离
$\because$直线上任意一点到原点的距离都大于等于原点到直线的距离
$\therefore \sqrt{a^2 + b^2} \geq \frac{\left| e^{\frac{x_0}{2}} \right|}{\sqrt{1 + 4x_0}}$
$\therefore a^2 + b^2 \geq \frac{e^{x_0}}{1 + 4x_0}$
剩下的过程同思路一