# Using Cauchy-Schwarz to prove inequality

I am trying to prove that for all $$x_1,\dots,x_n,y_1,\dots,y_n \in \mathbb R$$

$$\left( \sum_{i=1}^{n}x_i\ y_i \right)^2 \leq \ \left(\sum_{i=1}^{n}i \ x_i^2 \right) \cdot \left( \sum_{i=1}^{n} \frac{y_i^2}{i} \right)$$

I know the cauchy-schwarz inequality is $$(\sum_{i=1}^{n}x_i\ y_i)^2 \leq \ (\sum_{i=1}^{n}x_i^2) \cdot (\sum_{i=1}^{n} y_i^2)$$

I am trying to show that $$(\sum_{i=1}^{n}x_i^2) \cdot (\sum_{i=1}^{n} y_i^2) \leq \ (\sum_{i=1}^{n}i \ x_i^2) \cdot (\sum_{i=1}^{n} \frac{y_i^2}{i})$$ but I don't know how to continue from there..

• well, $x_iy_i = \tilde x_i \tilde y_i$ with $\tilde x_i = \sqrt ix_i$ and $\tilde y_i = \frac{y_i}{\sqrt i}$... so just use CS on $\tilde x, \tilde y$
– Surb
Jun 9 '17 at 15:16

$\displaystyle\left(\sum_{i=1}^nx_iy_i\right)^2=\left(\sum_{i=1}^n\left(\sqrt ix_i\right)\left(\frac{y_i}{\sqrt i}\right)\right)^2\leqslant\left(\sum_{i=1}^ni{x_i}^2\right)\left(\sum_{i=1}^n\frac{{y_i}^2}i\right)$
Hint: Write $$\left(\sum_{i=1}^n x_iy_i\right)^2 = \left(\sum_{i=1}^n (x_i \sqrt{i}) \left( \frac{y_i}{\sqrt{i}}\right)\right)^2.$$ What happens when you apply CS to this?