The Cauchy-Schwarz Inequality (which is known by other names, including Cauchy's Inequality, Schwarz's Inequality, and the Cauchy-Bunyakovsky-Schwarz Inequality) is a well-known inequality with many elegant applications. It has an elementary form, a complex form, and a general form.
Louis Cauchy wrote the first paper about the elementary form in 1821. The general form was discovered by Bunyakovsky in 1849 and independently by Schwarz in 1888.
Elementary Form
For any real numbers
and
,
with equality when there exists a nonzero constant
such that for all
,
.


![\[\left( \sum_{i=1}^{n}a_ib_i \right)^2 \le \left(\sum_{i=1}^{n}a_i^2 \right) \left(\sum_{i=1}^{n}b_i^2 \right)\]](https://latex.artofproblemsolving.com/b/2/b/b2b361a4b4b94a732f3b4cdd018274cb21ac53b1.png)



Discussion
Consider the vectors
and
. If
is the angle formed by
and
, then the left-hand side of the inequality is equal to the square of the dot product of
and
, or
.The right hand side of the inequality is equal to
. The inequality then follows from
, with equality when one of
is a multiple of the other, as desired.











Complex Form
The inequality sometimes appears in the following form.
Upper Bound on (Σa)(Σb)
Let
and
be two sequences of positive real numbers with
for
. Then
with equality if and only if, for some ordering of the pairs
, some
exists such that
for
and
for
, and
If we restrict that
and
for all
, then it's clear that for
to be
or
for all
, then
and
, so
is equivalent to
(When this is not an integer, the maximum occurs when
is either the ceiling or floor of the right-hand side.) In the special case that
is constant for all
, we have
and
, so here
must be
.


![\[0 < m \le \frac{a_i}{b_i} \le M\]](https://latex.artofproblemsolving.com/4/7/9/479dafe363dff8a8292852e8e7baf2c9a86c9a4b.png)

![\[\left(\sum_{i=1}^{n}a_i^2 \right) \left(\sum_{i=1}^{n}b_i^2 \right) \le \frac{(M+m)^2}{4Mm} \left( \sum_{i=1}^{n}a_ib_i \right)^2,\]](https://latex.artofproblemsolving.com/d/7/8/d7862cd2a62ebed9b7a5224be056b5ac8804efb9.png)






![\[m\sum_{\sigma(i)=1}^{j}b_{\sigma(i)}^2 = M\sum_{\sigma(i)=j+1}^{n}b_{\sigma(i)}^2.\]](https://latex.artofproblemsolving.com/c/c/8/cc8f8bebbd5bed4bb3de050ba118ce2c12057e52.png)









![\[m\sum_{\sigma(i)=1}^{j}b_{\sigma(i)}^2 = M\sum_{\sigma(i)=j+1}^{n}b_{\sigma(i)}^2\]](https://latex.artofproblemsolving.com/8/a/c/8acd2962fd7d4466773b4cdc155b70808df01406.png)








Proof
Note that for all
, we have
or
with equality if and only if
or
. Summing up these inequalities over
, we obtain from AM-GM that
and squaring gives us the desired bound. For equality to occur, we must have
or
for all
. If, without loss of generality,
for
and
for
for some
, then for the AM-GM to reach equality we must have (assume
since
is trivial)

![\[0 \le \left(\frac{a_i}{b_i}-m\right)\left(M-\frac{a_i}{b_i}\right) = \frac{1}{b_i^2}(a_ib_iM-a_i^2-b_i^2Mm+a_ib_im)\]](https://latex.artofproblemsolving.com/b/3/0/b304a3e522a3128582dfd84be56b841353c16fdc.png)
![\[(M+m)a_ib_i \ge a_i^2+(Mm)b_i^2,\]](https://latex.artofproblemsolving.com/9/4/8/948860f982853368e11e4b8124ffbb684e34d310.png)















General Form
Let
be a vector space, and let
be an inner product. Then for any
,
with equality if and only if there exist constants
not both zero such that
.



![\[\langle \mathbf{a,b} \rangle^2 \le \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle ,\]](https://latex.artofproblemsolving.com/b/2/a/b2a4377e6073ac3647d882b9d7c8a03dff4231d1.png)


Proof 1
Consider the polynomial of 
This must always be greater than or equal to zero, so it must have a non-positive discriminant, i.e.,
must be less than or equal to
, with equality when
or when there exists some scalar
such that
, as desired.

![\[\langle t\mathbf{a + b}, t\mathbf{a + b} \rangle = t^2\langle \mathbf{a,a} \rangle + 2t\langle \mathbf{a,b} \rangle + \langle \mathbf{b,b} \rangle .\]](https://latex.artofproblemsolving.com/5/a/d/5ad47b942f91f82081dfa331af5c158ba0bb93d4.png)





Proof 2
We consider
Since this is always greater than or equal to zero, we have
Now, if either
or
is equal to
, then
. Otherwise, we may normalize so that
, and we have
with equality when
and
may be scaled to each other, as desired.
![\[\langle \mathbf{a-b, a-b} \rangle = \langle \mathbf{a,a} \rangle + \langle \mathbf{b,b} \rangle - 2 \langle \mathbf{a,b} \rangle .\]](https://latex.artofproblemsolving.com/2/0/5/2052d373567e65a87a61e781574ecee7ddc4838a.png)
![\[\langle \mathbf{a,b} \rangle \le \frac{1}{2} \langle \mathbf{a,a} \rangle + \frac{1}{2} \langle \mathbf{b,b} \rangle .\]](https://latex.artofproblemsolving.com/f/6/1/f61276a891f1804aeae72f38b7579758b2ebf5cc.png)





![\[\langle \mathbf{a,b} \rangle \le 1 = \langle \mathbf{a,a} \rangle^{1/2} \langle \mathbf{b,b} \rangle^{1/2} ,\]](https://latex.artofproblemsolving.com/8/3/2/832616c9a006605d2f810252c5d0c2439a8abcb2.png)


Examples
The elementary form of the Cauchy-Schwarz inequality is a special case of the general form, as is the Cauchy-Schwarz Inequality for Integrals: for integrable functions
,
with equality when there exist constants
not both equal to zero such that for
,![\[\mu \int_a^t f(x)dx = \lambda \int_a^t g(x)dx .\]](https://latex.artofproblemsolving.com/6/d/a/6daaa774dc17841bd47984a42a4d380327d42e4a.png)
![$f,g : [a,b] \mapsto \mathbb{R}$](https://latex.artofproblemsolving.com/7/8/2/7826e7076f5f70e629cbec249b03d63d35175f42.png)
![\[\biggl( \int_{a}^b f(x)g(x)dx \biggr)^2 \le \int_{a}^b \bigl[ f(x) \bigr]^2dx \cdot \int_a^b \bigl[ g(x) \bigr]^2 dx\]](https://latex.artofproblemsolving.com/8/3/b/83b272d33297a3c8484d8fe4a9b38f094cc7e779.png)

![$t \in [a,b]$](https://latex.artofproblemsolving.com/5/1/b/51bf094096de5662cfd735ae3b56a9b87564cd56.png)
![\[\mu \int_a^t f(x)dx = \lambda \int_a^t g(x)dx .\]](https://latex.artofproblemsolving.com/6/d/a/6daaa774dc17841bd47984a42a4d380327d42e4a.png)
Problems
Introductory
- Consider the function
, where
is a positive integer. Show that
. (Source)
- (APMO 1991 #3) Let
,
,
,
,
,
,
,
be positive real numbers such that
. Show that
![\[\frac {a_1^2}{a_1 + b_1} + \frac {a_2^2}{a_2 + b_2} + \cdots + \frac {a_n^2}{a_n + b_n} \geq \frac {a_1 + a_2 + \cdots + a_n}{2}\]](https://latex.artofproblemsolving.com/e/c/f/ecfbc0cb313504e1ef9bc081535daeba1344325e.png)
Intermediate
- Let
be a triangle such that
![\[\left( \cot \frac{A}{2} \right)^2 + \left( 2 \cot \frac{B}{2} \right)^2 + \left( 3 \cot \frac{C}{2} \right)^2 = \left( \frac{6s}{7r} \right)^2 ,\]](https://latex.artofproblemsolving.com/4/c/7/4c7851c964aa97f92c013fa6e20faf6857a15495.png)




Olympiad
is a point inside a given triangle
.
are the feet of the perpendiculars from
to the lines
, respectively. Find all
for which
![\[\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}\]](https://latex.artofproblemsolving.com/b/b/6/bb6cdddd2246d364d5283b5e9c6c5922adfe433e.png)
AoPS
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