# FA5

Let $x_j,y_j\isin\mathbb{R}\,$ for $j=1,2,...,n\,$.

## (i)

Show that $\sum_{j=1}^n x_j y_j \le \left( \sum_{j=1}^n x_j^2 \right)^{1/2} \left( \sum_{j=1}^n y_j^2 \right)^{1/2}\,$.

Let $f(t)\,$ be a quadratic function $f(t) = t^2 \sum_{j=1}^n x_j^2 + 2t \sum_{j=1}^n x_jy_j + \sum_{j=1}^n y_j^2\,$.

$f(t) = \sum_{j=1}^n (t^2x_j^2 + 2tx_jy_j + y_j^2)\,$

$= \sum_{j=1}^n (tx_j+y_j)^2 \ge 0\,$ for $t,x_j,y_j\isin\mathbb{R}\,$.

Since the parabola $f(t):\mathbb{R}\to\mathbb{R}\,$ has at most one zero, its discriminant is less than or equal to zero.

$4\left( \sum_{j=1}^n x_jy_j \right)^2 - 4\sum_{j=1}^n x_j^2 \cdot \sum_{j=1}^n y_j^2 \le 0\,$

And therefore

$\sum_{j=1}^n x_jy_j \le |\sum_{j=1}^n x_jy_j| \le \left( \sum_{j=1}^n x_j^2 \right)^{1/2} \left(\sum_{j=1}^n y_j^2\right)^{1/2}\,$

## (ii)

Use (i) to prove that $\sum_{j=1}^n|x_jy_j| \le \left( \sum_{j=1}^n|x_j|^2\right)^{1/2}\left( \sum_{j=1}^n|y_j|^2\right)^{1/2}\,$

In the result (i), replace xj by | xj | and yj by | yj | .

$\sum_{j=1}^n |x_jy_j| = \sum_{j=1}^n |x_j||y_j| \le \left( \sum_{j=1}^n |x_j|^2 \right)^{1/2} \left(\sum_{j=1}^n |y_j|^2\right)^{1/2}\,$

## (iii)

Use (ii) to get $\left( \sum_{j=1}^n|x_j+y_j|^2 \right)^{1/2} \le \left( \sum_{j=1}^n|x_j|^2\right)^{1/2}+\left( \sum_{j=1}^n|y_j|^2\right)^{1/2}\,$

Notice $\sum_{j=1}^n |x_j + y_j|^2 \le \sum_{j=1}^n \left( |x_j|+|y_j| \right)^2\,$ by the triangle inequality.

$=\sum_{j=1}^n |x_j|^2 + 2\sum_{j=1}^n |x_jy_j| + \sum_{j=1}^n|y_j|^2\,$

$\le \sum_{j=1}^n |x_j|^2 + 2\left( \sum_{j=1}^n |x_j|^2 \right)^{1/2}\left( \sum_{j=1}^n |y_j|^2 \right)^{1/2} + \sum_{j=1}^n |y_j|^2\,$ from (ii).

$=\left[\left( \sum_{j=1}^n |x_j|^2 \right)^{1/2} + \left(\sum_{j=1}^n |y_j|^2\right)^{1/2}\right]^2\,$

Since $\sum_{j=1}^n |x_j+y_j|^2\ge 0\,$, we conclude,

$\left(\sum_{j=1}^n |x_j+y_j|^2\right)^{1/2} \le \left(\sum_{j=1}^n |x_j|^2\right)^{1/2} + \left( \sum_{j=1}^n |y_j|^2 \right)^{1/2}\,$.

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