<s>[INST] Maximum value of expression: $\frac{ab+bc+cd}{a^2+b^2+c^2+d^2}$ What is the maximum value of 
$$\frac{ab+bc+cd}{a^2+b^2+c^2+d^2}$$ 
where $a,b,c$, and $d$ are real numbers?
 [/INST] The main goal of my approach to the solution was to bring out an equality of the form $ab+bc+cd \le k(a^2+b^2+c^2+d^2)$, and thus, having our maximum value of this expression to be $k$
An initial attempt at the AM-GM inequality yields:
$\frac{1}{2}a^2+b^2+c^2+\frac{1}{2}d^2 \ge ab+bc+cd$
This result, as it turns out, isn't very useful.
Next I try splitting the expression uniformly:
$a^2+b^2+c^2+d^2 = a^2+mb^2+(1-m)b^2+mc^2+(1-m)c^2+d^2$
It is important to note that $m$ is a real number within $(0,1)$, so that both $mb^2,(1-m)b^2$ are positive, as the AM-GM inequality holds good only for positive real numbers.
Now, I individually apply the AM-GM inequality:
$a^2+mb^2 \ge 2\sqrt{m}ab$
$(1-m)b^2+(1-m)c^2 \ge 2(1-m)ab$
$mc^2+d^2 \ge 2\sqrt{m}cd$
Now, I take a value $m$ such that $\sqrt{m} = (1-m)$
We have: $m^2-3m+1 = 0$
$$m = \frac{3 \pm \sqrt{5}}{2}$$
But since, $m < 1$, we have: $m = \frac{3-\sqrt{5}}{2}$
Thus, $2\sqrt{m}=2(1-m)=\sqrt{5}-1$
Thus,we have:
$$a^2+b^2+c^2+d^2 \ge (\sqrt{5}-1)(ab+bc+cd)$$
So, to conclude:
$$\frac{ab+bc+cd}{a^2+b^2+c^2+d^2} \le \frac{1}{\sqrt{5}-1} = \frac{\sqrt{5}+1}{4}$$
The maximum value is $\frac{\sqrt{5}+1}{4}$
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