# Proof

(related to Proposition: Cauchy–Schwarz Inequality)

• By hypothesis, $x_1,x_2,\ldots x_n$ and $y_1,y_2,\ldots y_n$ are real numbers.
• The estimation is trivial, if all of them equal $0$ therefore, assume that both square roots $$A:=\left(\sum _{i=1}^{n}x_{i}^{2}\right)^{\frac 12},\quad B:=\left(\sum _{i=1}^{n}y_{i}^{2}\right)^{\frac 12}$$ are positive.
• Set $a_i:=\frac{|x_i|}A$, $b_i:=\frac{|y_i|}B,$ where $|x_i|$ and $|y_i|$ denote the respective absolute values of $x_i$ and $y_i$ for $i=1,\ldots,n.$
• Dividing by $AB$ and applying the 6th rule of calculations with inequalities the inequality $$\sum _{i=1}^{n}|x_{i}y_{i}|\leq \left(\sum _{i=1}^{n}x_{i}^{2}\right)^{\frac 12}\left(\sum _{i=1}^{n}y_{i}^{2}\right)^{\frac 12}$$ is equivalent to $$\sum _{i=1}^{n}a_{i}b_{i}=\sum _{i=1}^{n}\frac{|x_{i}|}A\frac{|y_{i}|}B\leq 1.$$
• Since by definition $a_ib_i > 0$, we have $a_ib_i=\sqrt{a_i^2b_i^2}.$
• Because $0\le \left(\sqrt{a_i^2}-\sqrt{b_i^2}\right)^2=a_i^2-2\sqrt{a_i^2b_i^2}+b_i^2$ for all $i=1,\ldots,n$ we have $$\sqrt{a_i^2b_i^2}\le \frac{a_i^2}2+\frac{b_i^2}2,\quad i=1,\ldots,n.$$
• Therefore, we have $$\sum _{i=1}^{n}a_{i}b_{i}\le \sum _{i=1}^{n}\left(\frac{a_i^2}2+\frac{b_i^2}2\right)=\frac 12\left(\sum _{i=1}^{n} a_i^2+\sum _{i=1}^{n}b_i^2\right)=\frac 12(1+1) = 1.$$

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### References

#### Bibliography

1. Heuser Harro: "Lehrbuch der Analysis, Teil 1", B.G. Teubner Stuttgart, 1994, 11th Edition