Cauchy's Inequality/Proof 2

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Theorem

$\ds \sum {r_i^2} \sum {s_i^2} \ge \paren {\sum {r_i s_i} }^2$


Proof

From the Complex Number form of the Cauchy-Schwarz Inequality, we have:

$\displaystyle \sum \left|{w_i}\right|^2 \left|{z_i}\right|^2 \ge \left|{\sum w_i z_i}\right|^2$

where all of $w_i, z_i \in \C$.


As elements of $\R$ are also elements of $\C$, it follows that:

$\displaystyle \sum \left|{r_i}\right|^2 \left|{s_i}\right|^2 \ge \left|{\sum r_i s_i}\right|^2$

where all of $r_i, s_i \in \R$.


But from the definition of modulus, it follows that:

$\displaystyle \forall r_i \in \R: \left|{r_i}\right|^2 = r_i^2$


Thus:

$\displaystyle \sum {r_i^2} \sum {s_i^2} \ge \left({\sum {r_i s_i}}\right)^2$

where all of $r_i, s_i \in \R$.

$\blacksquare$


Source of Name

This entry was named for Augustin Louis Cauchy.