Prove something similar to a variant of Cauchy-Shwarz inequality Cauchy-Shwarz Inequality is:
$$(a_1b_1 + a_2b_2 + \cdots + a_nb_n)^2 \leq (a_1^2 + a_2^2 + \cdots + a_n^2)(b_1^2 + \cdots + b_n^2)$$
However, it can be manipulated as:
$$\sqrt{(a_1-b_1)^2 + (a_2-b_2)^2 + \cdots + (a_n-b_n)^2} \leq \sqrt{a_1^2 + a_2^2 + \cdots + a_n^2} + \sqrt{b_1^2 + b_2^2 + \cdots + b_n^2}$$
I'm tasked with proving the following inequality:
$$\sqrt{(a_1 + b_1 + \cdots + z_1)^2 + (a_2 + b_2 + \cdots + z_2)^2 + \cdots + (a_n + b_n \cdots + z_n)^2} \leq \sqrt{a_1^2 + a_2^2 + \cdots + a_n^2} + \sqrt{b_1^2 + b_2^2 + \cdots + b_n^2} + \cdots + \sqrt{z_1^2 + z_2^2 + \cdots + z_n^2}$$
I've proved both Cauchy-Shwarz and its manipulation, but am lost when it come to the inequality right above. Hints and/or solutions are welcome.

Hint:
Replacing $b_i$ by $-b_i$ one can transform
$$
\sqrt{(a_1-b_1)^2+\cdots+(a_n-b_n)^2}\leq \sqrt{a_1^2+\cdots+a_n^2}+\sqrt{b_1^2+\cdots+b_n^2}
$$
into 
$$
\sqrt{(a_1+b_1)^2+\cdots+(a_n+b_n)^2}\leq \sqrt{a_1^2+\cdots+a_n^2}+\sqrt{b_1^2+\cdots+b_n^2}
$$
Then,
\begin{align*}
\sqrt{(a_1+b_1+c_1)^2+\cdots+(a_n+b_n+c_n)^2}&=\sqrt{(a_1+(b_1+c_1))^2+\cdots+(a_n+(b_n+c_n))^2}\\
&\leq\sqrt{a_1^2+\cdots+a_n^2}+\sqrt{(b_1+c_1)^2+\cdots+(b_n+c_n)^2}.
\end{align*}
Applying the result again, to the second radical, we get
\begin{align*}
\sqrt{a_1^2+\cdots+a_n^2}&+\sqrt{(b_1+c_1)^2+\cdots+(b_n+c_n)^2}\\
&\leq \sqrt{a_1^2+\cdots+a_n^2}+\sqrt{b_1^2+\cdots+b_n^2}+\sqrt{c_1^2+\cdots+c_n^2}.
\end{align*}
Now, use induction to complete the proof.
