prove that $\left\lvert \sum_{i=1}^{n} a_i \right\lvert \ ≤ \ \sum_{i=1}^n \lvert a_i \lvert$

Question

Please comment on my proof, see if I made some mistakes.

For $n \in \mathbb{N^+}$, $a_i \in \mathbb{R} (1 ≤ i ≤ n)$, prove that

$$\left\lvert \sum_{i=1}^{n} a_i \right\lvert \ ≤ \ \sum_{i=1}^n \lvert a_i \lvert$$

Denote the above proposition as $p(n)$, then let's proceed by induction.

Base case

$$p(2)\implies \left\lvert \sum_{i=1}^{n} a_i \right\lvert \ ≤ \ \sum_{i=1}^n \lvert a_i \lvert \\ \lvert a_1+a_2\lvert≤|a_1|+|a_2|$$

According to the triangular inequality $p(2)$ is true.

By induction Let's assume that $p(k)$ is true, and $k\in \mathbb{Z}$, then $p(k) \implies p(k+1)$ $$p(k)\implies \left\lvert \sum_{i=1}^{k} a_i \right\lvert \ ≤ \ \sum_{i=1}^k \lvert a_i \lvert \\ \\ \text{by the triangular inequality} \\ p(k+1)\implies \left\lvert \sum_{i=1}^{k} a_i + a_{k+1}\right\lvert \ ≤ \ \left\lvert \sum_{i=1}^k a_i \right\lvert + |a_{k+1}| \\ \text{but then} \left\lvert \sum_{i=1}^{k} a_i \right\lvert + | a_{k+1}| \ ≤ \ \sum_{i=1}^k |a_i \lvert + |a_{k+1}| \\ \left\lvert \sum_{i=1}^{k+1} a_i \right\lvert \ ≤ \ \sum_{i=1}^{k+1} \lvert a_i \lvert $$

Therefore, by induction, $p(n)$ is true.

Answer 1

This is another proof by using Jensen inequality which states that for a random variable $X$ and a convex function $g(x)$,

$$E(g(X)) \geq g(E(X))$$

Now, define a discrete random variable $X$ with sample space $\left\{a_1, a_2, \ldots, a_n\right\}$ such that for all $i$, $P(X=a_i)= p_{a_i}= \frac{1}{n}$. Let the convex function $g(x) = |x|$, then,

$$E(g(X)) = \sum_{i=1}^{n}g(a_i)p_{a_i} = \frac{1}{n}\sum_{i=1}^{n}|a_i|$$

$$g(E(X)) = g\left(\sum_{i=1}^{n}a_ip_{a_i}\right) = \left|\frac{1}{n}\sum_{i=1}^{n}a_i\right| = \frac{1}{n}\left|\sum_{i=1}^{n}a_i\right|$$

Now, using Jensen inequality,

$$E(g(X)) \geq g(E(X)) \implies \frac{1}{n}\sum_{i=1}^{n}|a_i| \geq \frac{1}{n}\left|\sum_{i=1}^{n}a_i\right|$$