Prove $\left|\sum_{i=1}^n x_i y_i \right| \le \dfrac{1}{a} \sum_{i=1}^n {x_i}^2 + \dfrac{a}{4}\sum_{i=1}^n {y_i}^2$

Question

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

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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!

Answer 1

$$\left|\sum_{i=1}^n x_i y_i \right| \le \sum_{i=1}^n\left| x_i y_i \right| $$

And for each $i$

$$|x_iy_i| \le \frac{1}{a}x_i^2 + \frac{a}{4}y_i^2$$

because $x + y \geq 2\sqrt{xy}$ when $x, y \geq 0$

Answer 2

You could start with the other end, use AM-GM: $$\dfrac{1}{a} \sum_{i=1}^n {x_i}^2 + \dfrac{a}{4}\sum_{i=1}^n {y_i}^2 \ge \left( \sum_{i=1}^n {x_i}^2 \right)^{1/2} \left( \sum_{i=1}^n {y_i}^2 \right)^{1/2}$$ and finish off as you did with Cauchy-Schwarz.

Answer 3

Hint:

For $x,y\in\mathbb{R}$ and $a>0$ we have \begin{align*} 0&\leq\left(\dfrac{x}{\sqrt{a}}\pm\dfrac{\sqrt{a}y}{2}\right)^2\\ \text{i.e.}\qquad0&\leq\dfrac{1}{a}x^2\pm xy+\dfrac{a}{4}y^2 \\ |xy| &\leq\dfrac{1}{a}x^2+\dfrac{a}{4}y^2 \end{align*}

Answer 4

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