The proof goes something like this. Find a maximum value for $$\frac{ab+bc+cd}{a^2+b^2+c^2+d^2}$$ where $$a,b,c,d \in \mathbb{R}$$ We can instead find a maximum value $2k$ such that $$a^2+b^2+c^2+d^2 \geq 2k(ab+bc+cd)$$ $$a^2+b^2+c^2+d^2 - 2k(ab+bc+cd) \geq 0$$ Let vector $v = (a, b, c, d)$. Then write the function as $$vM_kv^T \geq 0$$ Where T is the transpose and $$ M_k = \begin{bmatrix}1 & -k& 0 & 0\\ -k & 1& -k & 0\\ 0& -k& 1& -k\\ 0& 0& -k& 1\end{bmatrix}$$ Which is equivalent to setting the minimum eigenvalue of $M_k$ to be -non-negative. Solving for the characteristic polynomial, $$min(\lambda_k)= 1-k\sqrt{\frac{3 + \sqrt{5}}{2}} \geq 0$$ Therefore, $$2k \leq \frac{4}{\sqrt{5}+1}$$ $$\Rightarrow \frac{ab+bc+cd}{a^2+b^2+c^2+d^2} \leq \frac{1}{2k}$$ $$max(\frac{ab+bc+cd}{a^2+b^2+c^2+d^2})=\frac{\sqrt{5}+1}{4}$$
I understand why we would substitute $2k$ as a variable to find the maximum and getting everything $\geq 0$. However, I do not understand why we are able to choose a vector and define it as $v = (a, b, c, d)$ and then translate this to a matrix to solve for the maximum. I especially do not understand how the minimum eigenvalue is related to the answer (at least not really). I understand that we need this matrix to be positive definite and symmetric so that $vMv^T$ is defined, so (I guess?) solving for $k$ to be greater than $0$ makes sense. After that, I understand the substitutions and algebra to get the max. I guess my main questions are:
1) How exactly is the matrix composed given the definition of the vector $v$? I do not at all understand the parallels between the function and the matrix. What am I missing?
2) I want to say I do understand how to find the characteristic polynomial and the minimum eigenvalue and what they mean generally (i.e. an eigenvalue is $Mv = \lambda v$, thus the linear transformation $M$ scales the corresponding eigenvector $v$ by $\lambda$, thus it remains on its span after the transformation). I do not at all understand how this relates to the function that composes the matrix, which is likely because I do not understand how the matrix is composed. Are we trying to find the minimum value $min(\lambda_k)$ that keeps $M_k$ orthogonal, so in turn really we are just finding a better grid for the matrix described by the function that retains its geometry, or something along those lines? How does any of this relate to the maximum of the function? How else could I use this process?
The answer is correct, but I cannot say that the argument is correct because I find it hard to understand.
Here is an easier approach. Since both the numerator and denominator are quadratic polynomials in $a,b,c,d$, you may assume that $a^2+b^2+c^2+d^2=1$. So, you are maximising $\frac12v^TAv$ subject to $\|v\|=1$, where $v=(a,b,c,d)^T$ and $$ A=\pmatrix{0&1&0&0\\ 1&0&1&0\\ 0&1&0&1\\ 0&0&1&0}. $$ It follows that the maximum occurs when $v$ is a unit eigenvector of $A$ corresponding to the maximum eigenvalue. One can verify that $\lambda=\frac{1+\sqrt{5}}{2}$ and $$ v =\pmatrix{\sin\frac{\pi}{5}\\ \sin\frac{2\pi}{5}\\ \sin\frac{3\pi}{5}\\ \sin\frac{4\pi}{5}} =\pmatrix{\frac{1}{\sqrt{5+\sqrt{5}}}\\ \frac{1}{\sqrt{5-\sqrt{5}}}\\ \frac{1}{\sqrt{5-\sqrt{5}}}\\ \frac{1}{\sqrt{5+\sqrt{5}}}} $$ is a corresponding unit eigenvector. Therefore, the maximum possible value of $\frac{ab+bc+cd}{a^2+b^2+c^2+d^2}$ is $\frac12v^TAv=\frac12\lambda=\frac{1+\sqrt{5}}{4}$.