11. (20 points) Given $a_{1}, a_{2}, a_{3}, a_{4}$ are positive real numbers,
$$
\sum_{i=1}^{4} i a_{i} \leqslant 10, a_{i} \geqslant \frac{1}{2^{4-i}}(i=1,2,3,4) \text {. }
$$

Let $f=\sum_{i=1}^{4} \frac{1}{1+a_{i}^{i}}$. Find the minimum value of $f$.
11. Notice,
$$
\begin{array}{l}
\left(a_{1}-1\right)^{2} \geqslant 0, a_{2}\left(a_{2}-1\right)^{2} \geqslant 0, \\
\left(a_{3}-1\right)^{2}\left(a_{3}^{2}+2\left(a_{3}-\frac{1}{2}\right)\left(a_{3}+1\right)\right) \geqslant 0, \\
\left(a_{4}-1\right)^{2}\left(2 a_{4}^{3}+a_{4}^{2}-1\right) \geqslant 0 .
\end{array}
$$

Rearranging, we get
$$
\begin{array}{l}
\frac{1}{1+a_{1}} \geqslant-\frac{1}{4} a_{1}+\frac{3}{4}, \frac{1}{1+a_{2}^{2}} \geqslant-\frac{1}{2} a_{2}+1, \\
\frac{1}{1+a_{3}^{3}} \geqslant-\frac{3}{4} a_{3}+\frac{5}{4}, \frac{1}{1+a_{4}^{4}} \geqslant-a_{4}+\frac{3}{2} . \\
\text { Therefore, } f \geqslant-\frac{1}{4}\left(a_{1}+2 a_{2}+3 a_{3}+4 a_{4}\right)+ \\
\frac{3}{4}+1+\frac{5}{4}+\frac{3}{2} \\
\geqslant 2,
\end{array}
$$

Equality holds if and only if \(a_{1}=a_{2}=a_{3}=a_{4}=1\).
Thus, the minimum value of \(f\) is 2.