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The following inequality can be proven as follows:

Let $n\geq3$ and $0=a_0<a_1<\dots<a_{n+1}$ such that $a_1a_2+a_2a_3+\dots+a_{n-1}a_n=a_na_{n+1}$. Show that \begin{equation*} \frac{1}{{a_3}^2-{a_0}^2}+\frac{1}{{a_4}^2-{a_1}^2}+\dots+\frac{1}{{a_{n+1}}^2-{a_{n-2}}^2}\geq\frac{1}{{a_{n-1}}^2}. \end{equation*}

Solution:

The expression on the left-hand side can be rewritten as $$ \frac{a_1^2 a_2^2}{a_1^2 a_2^2 a_3^2 - a_0^2 a_1^2 a_2^2} + \frac{a_2^2 a_3^2}{a_2^2 a_3^2 a_4^2 - a_1^2 a_2^2 a_3^2} + \cdots + \frac{a_{n-1}^2 a_n^2}{a_{n-1}^2 a_n^2 a_{n+1}^2 - a_{n-2}^2 a_{n-1}^2 a_n^2}. $$ Applying the Cauchy-Schwarz inequality then yields

$$ \begin{align*} &\frac{a_1^2 a_2^2}{a_1^2 a_2^2 a_3^2 - a_0^2 a_1^2 a_2^2} + \frac{a_2^2 a_3^2}{a_2^2 a_3^2 a_4^2 - a_1^2 a_2^2 a_3^2} + \cdots + \frac{a_{n-1}^2 a_n^2}{a_{n-1}^2 a_n^2 a_{n+1}^2 - a_{n-2}^2 a_{n-1}^2 a_n^2} \\ & \ge \frac{\left( a_1 a_2 + a_2 a_3 + \cdots + a_{n-1} a_n \right)^2}{a_1^2 a_2^2 a_3^2 - a_0^2 a_1^2 a_2^2 + a_2^2 a_3^2 a_4^2 - a_1^2 a_2^2 a_3^2 + \cdots + a_{n-1}^2 a_n^2 a_{n+1}^2 - a_{n-2}^2 a_{n-1}^2 a_n^2} \\ & = \frac{a_n^2 a_{n+1}^2}{a_{n-1}^2 a_n^2 a_{n+1}^2 - a_0^2 a_1^2 a_2^2} \ge \frac{1}{a_{n-1}^2}. \end{align*} $$

When does equality hold?

Daniel Fischer
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arabhi manachra
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  • $a_i\in\mathbb{R}$ or $a_i\in\mathbb{N}_0$? – Rammus Jun 14 '15 at 13:44
  • @Rammus - Uh, not specified – arabhi manachra Jun 14 '15 at 13:45
  • When is Cauchy-Schwarz inequality an equality? – Bernard Jun 14 '15 at 13:45
  • @Bernard When each $y_i$ is a scalar multiple of $x_i$, or each $x_i$ is a scalar multiple of $y_i$. But it is not clear to me how to break it up here. – arabhi manachra Jun 14 '15 at 13:52
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    Equality holds iff $$\displaystyle{\frac{\frac{a_1^2 a_2^2}{a_1^2 a_2^2 a_3^2 - a_0^2 a_1^2 a_2^2}}{a_1^2 a_2^2 a_3^2 - a_0^2 a_1^2 a_2^2}=\frac{\frac{a_2^2 a_3^2}{a_2^2 a_3^2 a_4^2 - a_1^2 a_2^2 a_3^2}}{a_2^2 a_3^2 a_4^2 - a_1^2 a_2^2 a_3^2}}=\cdots=\frac{\frac{a_{n-1}^2 a_n^2}{a_{n-1}^2 a_n^2 a_{n+1}^2 - a_{n-2}^2 a_{n-1}^2 a_n^2}}{a_{n-1}^2 a_n^2 a_{n+1}^2 - a_{n-2}^2 a_{n-1}^2 a_n^2}$$ – user26486 Jun 14 '15 at 13:53
  • $$\displaystyle{\left(x_1^2+x_2^2+\cdots x_n^2\right)\left(y_1^2+y_2^2+\cdots +y_n^2\right)\ge \left(x_1y_1+x_2y_2+\cdots x_ny_n\right)}$$ has equality iff $\displaystyle{\frac{x_1}{y_1}=\frac{x_2}{y_2}=\cdots =\frac{x_n}{y_n}}$ – user26486 Jun 14 '15 at 13:56
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    I.e. When $a_1a_2(a_3^2-a_0^2)$ and similar cyclic terms are all equal to some (obviously positive) constant. – Macavity Jun 14 '15 at 14:00
  • If there's no absolute value in your formula, you must add the common ratio is positive. – Bernard Jun 14 '15 at 14:01
  • @Bernard the equality condition comes from expressions such as $(\frac{a}x-\frac{b}y)^2\ge0$. Why can't the "common ratio" be negative throughout? In this problem the ratio is by definition positive, but can't see why we need that in general. – Macavity Jun 14 '15 at 16:44
  • I didn't examine the details of the computation. But in general,, equality in Cauchy-Schwarz without absolute value for the inner product requires colinearity and same direction of the vectors. Although there are several variants in formulation… – Bernard Jun 14 '15 at 16:52

1 Answers1

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Deduced from the equality condition of Cauchy-Schwarz inequality, we have: $$ \forall 0 \leqslant k \leqslant n-1, \, \, a_{k+3}^2 = a_k^2 + c$$ where $c > 0$ is a constant. So the equation has 3 possiblities that depends on $n$.

For example, if $n = 3K + 1$:

$$\sum_{k=0}^{K-1} (a_0 + kc)(a_1 + kc) + (a_1 + kc)(a_2 + kc) + (a_2 + kc)(a_0 + (k+1)c) = (a_1 + Kc)(a_2 + Kc) - (a_0 + Kc)(a_1 + Kc).$$

We can solve the above equation with respect to $c$(quadratic) and see what the roots are. If one of the roots is positive then the equality holds.

corindo
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