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.