Cauchy-Schwarz Inequality

Revision as of 01:26, 18 June 2006 by MCrawford (talk | contribs) (bolded article name)

The Cauchy-Schwarz Inequality (which is known by other names, including Cauchy's Inequality) states that, for two sets of real numbers $a_1,a_2,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$, the following inequality is always true:


Equality holds if and only if $\frac{a_1}{b_1}=\frac{a_2}{b_2}=...=\frac{a_n}{b_n}$.

There are many ways to prove this; one of the more well-known is to consider the equation $(a_1x+b_1)^2+(a_2x+b_2)^2+...+(a_nx+b_n)^2=0$. Expanding, we find the equation to be of the form $Ax^2+Bx+C$, where $A=\sum_{i=1}^n a_i^2$, $B=2\sum_{j=1}^n a_jb_j$, and $C=\sum_{k=1}^n b_k^2.$. By the Trivial Inequality, we know that the left-hand-side of the original equation is always at least 0, so either both roots are Complex Numbers, or there is a double root at $x=0$. Either way, the discriminant of the equation is nonpositive. Taking the discriminant, $B^2-4AC \leq 0$ and substituting the above values of A, B, and C leaves us with the Cauchy-Schwarz Inequality, $(a_1b_1+a_2b_2+...+a_nb_n)^2 \leq (a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...+b_n^2)$, or, in the more compact sigma notation, $\left(\sum a_ib_i\right)$ $\leq \left(\sum a_i^2\right)\left(\sum b_i^2\right)$

Note that this also gives us the equality case; equality holds if and only if the discriminant is equal to 0, which is true if and only if the equation has 0 as a double root, which is true if and only if $\frac{a_1}{b_1}=\frac{a_2}{b_2}=...=\frac{a_n}{b_n}$.

This inequality is used very frequently to solve Olympiad-level Inequality problems, such as those on the USAMO and IMO.

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