\(已知a>0,b>0,且a+b=1,求\frac{1}{4a}+\frac{4a}{2a+b}最小值。\)
\[
\begin{align*}
解：\frac{1}{4a}+\frac{4a}{2a+b} \\
& = \frac{a+b}{4a}+\frac{4a}{2a+b} \\
& = \frac{2a+b-a}{4a}+\frac{4a}{2a+b} \\
& = \frac{2a+b}{4a}+\frac{4a}{2a+b} -\frac{a}{4a} \\
& \geq 2-\frac{1}{4} \\
& \ge \frac{7}{4} \\
\end{align*}
\]