What is the maximum value of the expression?

Relevant wiki: Arithmetic Mean - Geometric Mean

We split $a^2+b^2+c^2+d^2$ as

$(a^2+ \lambda b^2) +((1-\lambda)b^2 + (1-\lambda)c^2)+(\lambda c^2 + d^2)$ , for some $\lambda$ in $(0,1)$ , and apply the AM-GM inequality to get

$a^{2} + b^{2} + c^{2} + d^{2} \geq 2\sqrt{\lambda} ab + 2(1-\lambda) bc + 2\sqrt{\lambda} cd$ . Now we choose a lambda such that $2\sqrt{\lambda} = 2(1-\lambda)$ , which simplifies to

$\lambda^{2} -3\lambda + 1 =0$ , and the root that lies in $(0,1)$ is $\lambda = \frac{3-\sqrt{5}}{2}$ . For this value , we get $2\sqrt{\lambda} = 2(1-\lambda) = \sqrt{5} -1$ , so

$\frac{ab+bc+cd}{a^2+b^2+c^2+d^2} \leq \frac{1}{\sqrt{5} -1} = \frac {\sqrt{5} +1}{4} = 0.809$ .

Equality holds iff $a= \sqrt{\lambda }b$ , $b=c$ and $a=d$ , i.e. $b=c=k$ and $a= d= \frac{(\sqrt{5}-1)k}{2}$ for any positive real $k$ .

Here is an alternative, linear-algebraic way to solve the problem, which does not involve any parameter tuning. Restated, the problem is to find the minimum value of $\frac{1}{2k}$ such that the following inequality holds, for all real $\mathbf{x}'=(a,b,c,d)$ , not all zero $a^2+b^2+c^2+d^2 \geq 2k(ab+bc+cd).$ Transposing, we need the quadratic form $\mathbf{x}' \mathbf{M(k)} \mathbf{x} \geq 0,$ where the matrix $\mathbf{M(k)}$ is given by $\mathbf{M}(k)=\begin{pmatrix} 1 & -k & 0 & 0 \\ -k & 1 & -k & 0 \\ 0 & -k & 1 & -k \\ 0 & 0 & -k & 1 \end{pmatrix}$ which is equivalent to requiring that its minimum eigenvalue be non-negative. Solving the characteristic equation, we readily obtain the minimum eigenvalue of $\mathbf{M}(k)$ as follows $\lambda_{\textrm{min}}(k)= 1- k\sqrt{\frac{3+\sqrt{5}}{2}},$ which is required to be non-negative. This yields, $\frac{1}{2k} \geq \frac{\sqrt{5}+1}{4},$ with the lower-bound clearly attainable.