The sum of the squares is less than or equal to the square of the sums for all n. I am trying to understand this proof. Rather an important part of the proof. I have already shown this is true for $n=2$ and am assuming the $a_n$ case is true.
$$(a_1^2+a_2^2+...+a_n^2)  \le (a_1+a_2+...+a_n)^2$$
Want to show that 
$$(a_1^2+a_2^2+...+a_n^2 + a_{n+1}^2)  \le (a_1+a_2+...+a_n+a_{n+1})^2$$
$=$
$$(a_1^2+a_2^2+...a_n^2) + a_{n+1}^2  \le ((a_1+a_2+...a_n)+(a_{n+1}))^2$$
$=$
$$(a_1^2+a_2^2+...+a_n^2 + a_{n+1}^2)  \le (a_1+a_2+...+a_n)^2+2(a_1+a_2+...+a_n)(a_{n+1})+(a_{n+1}) ^2$$ and here is the part I am not understanding. For some reason the proof moves some of the terms over and I cannot identify what is being replaced or why. My guess is that the terms that moves are the ${n+1}$ terms. But, I am not certain. 
$$a_1^2+a_n^2+a_{n+1}^2...+2(a_1+a_2+...a_n)(a_{n+1})  \le (a_1+a_2+...a_n)^2$$
 A: inductive step: 
the claim being correct for 
$$(a_1^2+a_2^2+...+a_n^2)  \le (a_1+a_2+...+a_n)^2$$
implies 
$$(a_1^2+a_2^2+...+a_n^2+a_{n+1}^2)  \le (a_1+a_2+...+a_n+a_{n+1})^2$$
Proof
\begin{align}
a_1^2+a_2^2+...+a_n^2+a_{n+1}^2 &=(a_1^2+a_2^2+...+a_n^2)+a_{n+1}^2\\
& \leq (a_1+a_2+...+a_n)^2+a_{n+1}^2 \mbox{ (using the assumption)}\\
&=y^2+a_{n+1}^2 \mbox{ (rewriting } y=a_1+a_2+...+a_n)\\
& \leq (y+a_{n+1})^2 \mbox{ (using: } a^2+b^2\leq (a+b)^2)\\
&=(a_1+a_2+...+a_n+a_{n+1})^2 \mbox{ (plug back for } y)
\end{align}
A: If you really need to use induction, here's what you need for the inductive step:
Assuming $(a_1^2+a_2^2+...+a_n^2 )  \le (a_1+a_2+...+a_n)^2$
then
$$\begin{align} (a_1+a_2+...+a_n+a_{n+1})^2 &= ((a_1+a_2+...a_n)+(a_{n+1}))^2 \\
&= (a_1+a_2+...+a_n)^2+2(a_1+a_2+...+a_n)a_{n+1} + a_{n+1}^2 \\
&\ge (a_1+a_2+...+a_n)^2 + a_{n+1}^2 \tag{*}\\
&\ge (a_1^2+a_2^2+...+a_n^2 ) + a_{n+1}^2 \\
\end{align}
$$as required. Note that $2(a_1+a_2+...+a_n)a_{n+1}\ge 0)$ for the (*) step.
You do require that all the $a_i$ are not negative.
Being visual myself, I prefer pictures....

