已知正实数 $x, y$ 满足 $x + 2y + \sqrt{xy} = 7$，求 $x + 3y$ 的最小值。
解：$\because x > 0,\ y > 0$
$\therefore ax + \frac{y}{a} \geq 2\sqrt{xy}$
$\therefore x + 2y + \sqrt{xy} \leq x + 2y + \frac{1}{2}\left(ax + \frac{y}{a}\right) = \left(1 + \frac{1}{2}a\right)x + \left(2 + \frac{1}{2a}\right)y$
$\therefore 7 \leq \left(1 + \frac{1}{2}a\right)x + \left(2 + \frac{1}{2a}\right)y$
$\text{令 }\frac{1 + \frac{1}{2}a}{2 + \frac{1}{2a}} = \frac{1}{3}\text{ 得 }a = \frac{1}{3}$
$\therefore 7 \leq \frac{7}{6}x + \frac{7}{2}y$
$\therefore x + 3y \geq 6$