- #61
Antarres
- 212
- 102
The solution to this inequality draws inspiration from this integral:
$$\int \frac{dx}{1+k^2x^2} = \frac{1}{k}\arctan(kx) + C$$
We assume that the sequences on the right-hand side are well defined. Then we won't have any convergence problems.
We proceed using Cauchy-Schwarz inequality:
$$\left(\sum_{n \in \mathbb{N}}a_n\right)^2 \leq \sum_{n \in \mathbb{N}} a^2_n(1+k^2n^2)\sum_{n \in \mathbb{N}} \frac{1}{1+k^2n^2} $$.
Here ##k \in \mathbb{R}## is just a parameter, setting it up for the integral above.
The second sum can be bounded by the integral:
$$\sum_{n \in \mathbb{N}} \frac{1}{1+k^2n^2} < \int_0^\infty \frac{dx}{1+k^2x^2} = \frac{\pi}{2k}$$.
Combining the inequalities, we have:
$$\left(\sum_{n \in \mathbb{N}}a_n\right)^2 < \sum_{n \in \mathbb{N}} a^2_n(1+k^2n^2)\frac{\pi}{2k} = \frac{\pi}{2} \left(\frac{1}{k}\sum_{n \in \mathbb{N}} a^2_n + k \sum_{n \in \mathbb{N}} n^2a^2_n\right)$$
The term on the right is of the form ##\tfrac{x}{k} + yk##, and we need it to be of the form ##2\sqrt{xy}## in order to prove the inequality. So we will assume that it's of this form, and search for ##k## to confirm it, if possible. We find:
$$\frac{x}{k} + yk = 2\sqrt{xy} \Rightarrow \frac{x^2}{k^2} - 2xy + y^2k^2 = 0 \Rightarrow \left(\frac{x}{k} - yk\right) = 0 \Rightarrow k = \sqrt{\frac{x}{y}}$$.
So with choice:
$$k = \sqrt{\frac{\sum_{n \in \mathbb{N}} a^2_n}{\sum_{n \in \mathbb{N}} n^2a^2_n}}$$
we obtain the inequality that we're looking for by squaring both sides.
$$\int \frac{dx}{1+k^2x^2} = \frac{1}{k}\arctan(kx) + C$$
We assume that the sequences on the right-hand side are well defined. Then we won't have any convergence problems.
We proceed using Cauchy-Schwarz inequality:
$$\left(\sum_{n \in \mathbb{N}}a_n\right)^2 \leq \sum_{n \in \mathbb{N}} a^2_n(1+k^2n^2)\sum_{n \in \mathbb{N}} \frac{1}{1+k^2n^2} $$.
Here ##k \in \mathbb{R}## is just a parameter, setting it up for the integral above.
The second sum can be bounded by the integral:
$$\sum_{n \in \mathbb{N}} \frac{1}{1+k^2n^2} < \int_0^\infty \frac{dx}{1+k^2x^2} = \frac{\pi}{2k}$$.
Combining the inequalities, we have:
$$\left(\sum_{n \in \mathbb{N}}a_n\right)^2 < \sum_{n \in \mathbb{N}} a^2_n(1+k^2n^2)\frac{\pi}{2k} = \frac{\pi}{2} \left(\frac{1}{k}\sum_{n \in \mathbb{N}} a^2_n + k \sum_{n \in \mathbb{N}} n^2a^2_n\right)$$
The term on the right is of the form ##\tfrac{x}{k} + yk##, and we need it to be of the form ##2\sqrt{xy}## in order to prove the inequality. So we will assume that it's of this form, and search for ##k## to confirm it, if possible. We find:
$$\frac{x}{k} + yk = 2\sqrt{xy} \Rightarrow \frac{x^2}{k^2} - 2xy + y^2k^2 = 0 \Rightarrow \left(\frac{x}{k} - yk\right) = 0 \Rightarrow k = \sqrt{\frac{x}{y}}$$.
So with choice:
$$k = \sqrt{\frac{\sum_{n \in \mathbb{N}} a^2_n}{\sum_{n \in \mathbb{N}} n^2a^2_n}}$$
we obtain the inequality that we're looking for by squaring both sides.