Q:

Help understanding example in Engel's *Problem Solving Strategies*

I've spent a lot of time trying to follow the chain of reasoning, but to no avail. I lose track of how it works at the "Adding (1) and (2)" part.
Could someone help me understand this, please?

A:

Okay, I figured it out (largely thanks to @Did's comment).
Here's the complete proof.
Consider the point $(a_n, b_n, c_n, d_n)$ in 4-space. Its distance from the origin is
$$
a_{n+1}^2 + b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2 = 2(a_n^2 + b_n^2 + c_n^2 + d_n^2) - 2(a_nb_n + b_nc_n +c_nd_n + a_nd_n) \;\;(1)
$$
Now, since $a_n + b_n + c_n + d_n \gt 0$ for all $n \gt 0$,
$$
\begin{align*}
0 \;
&= (a_{n} + b_{n} + c_{n} + d_{n})^2\\
&= (a_n + c_n)^2 + (b_n + d_n)^2 + 2(a_nb_n + b_nc_n +c_nd_n + a_nd_n)
 \end{align*}
$$
This implies
$$-2(a_nb_n + b_nc_n +c_nd_n + a_nd_n) = 
(a_n + c_n)^2 + (b_n + d_n)^2 \;\;\;\;(2)
$$
Now we can add $(1)$ and $(2)$ to get
$$ \begin{align*}
a_{n+1}^2 + b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2 &= 2(a_n^2 + b_n^2 + c_n^2 + d_n^2) + (a_n + c_n)^2 + (b_n + d_n)^2\\
&\ge 2(a_n^2 + b_n^2 + c_n^2 + d_n^2)\\
&\ge 2^2(a_{n-1}^2 + b_{n-1}^2 + c_{n-1}^2 + d_{n-1}^2)\\
&\ge \cdots\\
&\ge 2^{n-j}(a_{n-j}^2 + b_{n-j}^2 + c_{n-j}^2 + d_{n-j}^2)\\
&\ge \cdots\\
&\ge 2^n(a_1^2 + b_1^2 + c_1^2 + d_1^2)
 \end{align*} 
$$
Hence the distance of the point $(a_n, b_n, c_n, d_n)$ increases exponentially in 4-space, etc., etc., etc. $\square$
PS. If there are any errors, please tell me!

