\(已知:a+b=1,\)
\(求解:\frac{1}{4a}+\frac{4a}{2a+b}最小值。\)
\[ \because a+b=1 \] \[ \therefore \frac{1}{4a}+\frac{4a}{2a+b} = \frac{a+b}{4a}+\frac{4a}{2a+b} \\ \] \[ \begin{align*} & \frac{a+b}{4a}+\frac{4a}{2a+b} \\ & = \frac{a+b+a-a}{4a}+\frac{4a}{2a+b} \\ & = \frac{2a+b}{4a}-\frac{a}{4a}+\frac{4a}{2a+b} \\ & \geq 2\sqrt{\frac{2a+b}{4a} \cdot \frac{2a+b}{4a}} -\frac{1}{4} \\ & \geq 2-\frac{1}{4} \\ & \geq 1\frac{3}{4} \\ &\text{解得原式最小值为：} & 1\frac{3}{4} \end{align*} \]