Maximum value of expression: $\frac{ab+bc+cd}{a^2+b^2+c^2+d^2}$

Question

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?

Answer 1

Have a look at the Rayleigh quotient of two quadratic forms. This expression is homogeneous of degree zero in vector $(a,b,c,d)$ so you can look for its maximum on unit sphere.

Then if $X=(a,b,c,d)$ is on this sphere and $$Q = \frac{1}{2}\begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0\end{pmatrix}$$

You have

$$\frac{ab+bc+cd}{a^2+b^2+c^2+d^2} = \frac{X^T Q X}{X^TX}=X^T Q X$$

This is maximum for eigen-vector of $Q$ associated to its maximum eigen-value (which seems to be half of golden ratio $\frac{(1+\sqrt{5})}{4}$)

Answer 2

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 $\frac{\sqrt{5}+1}{4}$