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For any $a_j \in \Bbb R, \, j = 1, 2, \cdots, n$, one has the bound

$$\left | \sum_{j = 1}^n a_j \right |^2 + \left | \sum_{j = 1}^n (-1)^j a_j \right |^2 \le (n + 2) \sum_{j =1}^n a_j^2.$$

This is an exercise from Cauchy's inequality.

My first try:

From Cauchy's inequality we have

$$\left | \sum_{j = 1}^n a_j \right |^2 \le n \sum_{j = 1}^n a_j^2,$$

$$\left | \sum_{j = 1}^n (-1)^j a_j \right |^2 \le n \sum_{j = 1}^n a_j^2.$$

Add these two and we have

$$\left | \sum_{j = 1}^n a_j \right |^2 + \left | \sum_{j = 1}^n (-1)^j a_j \right |^2 \le 2n \sum_{j = 1}^n a_j^2,$$

which is weaker than the inequality in question.

My second try:

$$\left | \sum_{j = 1}^n a_j \right |^2 + \left | \sum_{j = 1}^n (-1)^j a_j \right |^2 = (\sum_{j \text{ is even}} a_{j} + \sum_{j \text{ is odd}} a_{j})^2 + (\sum_{j \text{ is even}} a_{j} - \sum_{j \text{ is odd}} a_{j})^2.$$

On the other hand,

\begin{align} (n + 2) \sum_{j = 1}^n a_j^2 & = n \sum_{j = 1}^n a_j^2 + 2 \sum_{j = 1}^n a_j^2 \\ & \ge (\sum_{j = 1}^n a_j)^2 + 2 \sum_{j = 1}^n a_j^2 \\ & = (\sum_{j \text{ is even}} a_{j} + \sum_{j \text{ is odd}} a_{j})^2 + 2 \sum_{j = 1}^n a_j^2. \\ \end{align}

So we only need to show that

$$(\sum_{j \text{ is even}} a_{j} + \sum_{j \text{ is odd}} a_{j})^2 + 2 \sum_{j = 1}^n a_j^2 \ge (\sum_{j \text{ is even}} a_{j} + \sum_{j \text{ is odd}} a_{j})^2 + (\sum_{j \text{ is even}} a_{j} - \sum_{j \text{ is odd}} a_{j})^2,$$

i.e.

$$2 \sum_{j = 1}^n a_j^2 \ge (\sum_{j = 1}^n (-1)^j a_j)^2.$$

Am I correct and how do I proceed?

Yuxiao Xie
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  • 2
    You can proceed from the first step of your second attempt as, \begin{align}&\left(\sum_{j \text{ is even}} a_{j} + \sum_{j \text{ is odd}} a_{j}\right)^2 + \left(\sum_{j \text{ is even}} a_{j} - \sum_{j \text{ is odd}} a_{j}\right)^2 \=& 2\left(\sum_{j \text{ is even}} a_{j}\right)^2 + 2\left(\sum_{j \text{ is odd}} a_{j}\right)^2\ \le & 2\left(\left[\frac{n}{2}\right]\right)\sum_{j \text{ is even}} a_{j}^2 + 2\left(\left[\frac{n}{2}\right]+1\right)\sum_{j \text{ is odd}} a_{j}^2\ \le & (n+2)\sum_{j=1}^{n} a_j^2\end{align} – r9m Apr 04 '16 at 07:25

1 Answers1

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Here is the proof given by my textbook:

$$LHS = 2 \sum_{j = 1}^n a_j^2 + 4 \sum_{(j, k) \in S} a_j a_k,$$

where $S$ is the set of all $(j, k)$ such that $1 \le j \lt k \le n$ with $j + k$ even.

\begin{align} LHS & \le 2 \sum_{j = 1}^n a_j^2 + 2 \sum_{(j, k) \in S} (a_j^2 + a_k^2) \\ & \le 2 \sum_{j = 1}^n a_j^2 + 2 \sum_{s = 1}^n n_s a_s^2, \\ \end{align}

where $n_s$ denotes the number of pairs $(j, k)$ in $S$ with $j = s$ or $k = s$.

Observe that

$$n_s \le \left \lfloor \frac {n - 1}2 \right \rfloor.$$

Thus,

\begin{align} LHS & \le (2 + 2 \left \lfloor \frac {n - 1}2 \right \rfloor) \sum_{j = 1}^n a_j^2 \\ & \le (n + 2) \sum_{j = 1}^n a_j^2. \\ \end{align}

Yuxiao Xie
  • 8,536