FA5

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Let x_{j},y_{j}\in {\mathbb  {R}}\, for j=1,2,...,n\,.


(i)

Show that \sum _{{j=1}}^{n}x_{j}y_{j}\leq \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_{j}y_{j}+\sum _{{j=1}}^{n}y_{j}^{2}\,.

f(t)=\sum _{{j=1}}^{n}(t^{2}x_{j}^{2}+2tx_{j}y_{j}+y_{j}^{2})\,

=\sum _{{j=1}}^{n}(tx_{j}+y_{j})^{2}\geq 0\, for t,x_{j},y_{j}\in {\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_{j}y_{j}\right)^{2}-4\sum _{{j=1}}^{n}x_{j}^{2}\cdot \sum _{{j=1}}^{n}y_{j}^{2}\leq 0\,


And therefore

\sum _{{j=1}}^{n}x_{j}y_{j}\leq |\sum _{{j=1}}^{n}x_{j}y_{j}|\leq \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_{j}y_{j}|\leq \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 x_{j} by |x_{j}| and y_{j} by |y_{j}|.


\sum _{{j=1}}^{n}|x_{j}y_{j}|=\sum _{{j=1}}^{n}|x_{j}||y_{j}|\leq \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}}\leq \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}\leq \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_{j}y_{j}|+\sum _{{j=1}}^{n}|y_{j}|^{2}\,

\leq \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}\geq 0\,, we conclude,


\left(\sum _{{j=1}}^{n}|x_{j}+y_{j}|^{2}\right)^{{1/2}}\leq \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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