Prove $\left|\sum_{i=1}^n x_i y_i \right| \le \dfrac{1}{a} \sum_{i=1}^n {x_i}^2 + \dfrac{a}{4}\sum_{i=1}^n {y_i}^2$ If $X,Y$ are vectors in $\mathbb{R}^n$ and $a>0$ show that:
$$\left|\sum_{i=1}^n x_i y_i \right| \le \dfrac{1}{a} \sum_{i=1}^n {x_i}^2 + \dfrac{a}{4}\sum_{i=1}^n {y_i}^2 (*)$$

I started with Cauchy–Schwarz inequality and got:
$$\left|\sum_{i=1}^n x_i y_i \right| \le  {\sum_{i=1}^n {x_i}^2}^\frac{1}{2}  \cdot{\sum_{i=1}^n {y_i}^2}^\frac{1}{2}(**)$$
So apparently we need to show that  $(**) < (*)$
and I'm stuck. Don't really know what to do with $a$'s.
Please help!

Observe that the inequality follows first from Cauchy-Schwarz inequality and then by AM-GM inequality.
