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?

  • $\begingroup$ I have a solution to this particular problem, and I'm posting it here. I'm trying to discover other methods, and I would welcome different ways of solving this problem. $\endgroup$
    – HPP_00
    Mar 18, 2017 at 12:03
  • $\begingroup$ The expression posed in the question's title differs from the one in the text. Namely the term $d^2$ in the denominator is missing in the expression in the text. $\endgroup$ Mar 18, 2017 at 13:35
  • $\begingroup$ @MichaelChernick Thank you. $\endgroup$
    – HPP_00
    Mar 18, 2017 at 16:42

2 Answers 2


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

  • $\begingroup$ I checked your computations by hand it works! This so so neat! But how did you find Q? $\endgroup$
    – Imago
    Mar 18, 2017 at 13:57
  • $\begingroup$ Btw. Typo: $cd$ instead of $bd$ $\endgroup$
    – Imago
    Mar 18, 2017 at 14:02
  • 1
    $\begingroup$ Oops that is correct, thank you. For any quadratic form $ax^2+by^2+cz^2+2dyz+2exz+2fxy$, there is a unique symmetric matrix $Q$ associated with this form such that it is equal to $X^T Q X$, this matrix being $$Q = \begin{pmatrix}a & f & e \\ f & b & d \\ e & d & c\end{pmatrix}$$ and where $X=(x,y,z)$. This can be extended to any dimension (and here in $\mathbb{R}^4$) $\endgroup$
    – M. Boyet
    Mar 18, 2017 at 14:08

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

  • $\begingroup$ Extremely elegant! Incidentally, a good way to arrive at the golden ratio. Upvote! $\endgroup$
    – GRrocks
    Mar 18, 2017 at 13:50

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