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## Main Question or Discussion Point

Hi everyone,

I am working on my own through Rudin's Principles of Mathematical Analysis and, after the demonstration of Cauchy - Schwarz Inequality, in Theorem 1.37, part (d), Rudin states:

$$|x \cdot y| \leqslant |x||y|$$

When he explains how to prove this, he simply states that this is an immediate consequence of Schwarz Inequality, which he defines as follows:

$$|\sum_{j=1}^{n}a_j \overline{b_j}|^2 \leqslant \sum_{j=1}^{n}|a_j|^2 \sum_{j=1}^{n}|b_j|^2$$

If someone can explain me how this two things are identical I would appreciate it a lot. My toughts so far:

$$|x \cdot y| \leqslant |x||y|$$ Take the square of this, which is:

$$ (x \cdot y)(x \cdot y) \leqslant \sum x_i^2 \sum y_i^2$$ and hence,

$$ (\sum x_iy_i)^2 \leqslant \sum x_i^2 \sum y_i^2$$ which is NOT the Schwarz Inequality!

I am working on my own through Rudin's Principles of Mathematical Analysis and, after the demonstration of Cauchy - Schwarz Inequality, in Theorem 1.37, part (d), Rudin states:

$$|x \cdot y| \leqslant |x||y|$$

When he explains how to prove this, he simply states that this is an immediate consequence of Schwarz Inequality, which he defines as follows:

$$|\sum_{j=1}^{n}a_j \overline{b_j}|^2 \leqslant \sum_{j=1}^{n}|a_j|^2 \sum_{j=1}^{n}|b_j|^2$$

If someone can explain me how this two things are identical I would appreciate it a lot. My toughts so far:

$$|x \cdot y| \leqslant |x||y|$$ Take the square of this, which is:

$$ (x \cdot y)(x \cdot y) \leqslant \sum x_i^2 \sum y_i^2$$ and hence,

$$ (\sum x_iy_i)^2 \leqslant \sum x_i^2 \sum y_i^2$$ which is NOT the Schwarz Inequality!