Set $x_i :=1/(1+a_i)$ and $y_i:=1/a_i$, and let $a_1\le a_2\le \cdots \le a_n$ without loss of generality. Observe that $y_i-x_i = x_i y_i$. Using this, it is equivalent to proving \[ \sum_{1\le i\le n}x_iy_i \ge \frac1n\left(\sum_{1\le i\le n}x_i\right)\left(\sum_{1\le i\le n}y_i\right) \Leftrightarrow \frac1n\sum_{1\le i\le n}x_iy_i \ge \left(\frac1n\sum_{1\le i\le n}x_i\right)\left(\frac1n\sum_{1\le i\le n}y_i\right). \]Since $x_1\ge \cdots \ge x_n$ and $y_1\ge \cdots \ge y_n$ by Chebyshev's sum inequality, the proof is complete.