Cauchy's Inequality/Proof 2

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Theorem

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

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


Proof

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

$\ds \paren {\sum \cmod {w_i}^2} \paren {\sum \cmod {z_i}^2} \ge \cmod {\sum w_i z_i}^2$

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


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

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

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


But from the definition of modulus, it follows that:

$\ds \forall r_i \in \R: \size {r_i}^2 = {r_i}^2$


Thus:

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

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

$\blacksquare$


Source of Name

This entry was named for Augustin Louis Cauchy.