Difference between revisions of "Cauchy-Schwarz Inequality"
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− | 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. | + | 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. |
+ | |||
+ | [https://en.wikipedia.org/wiki/Augustin-Louis_Cauchy|Augustin Louis Cauchy] wrote the first paper about the elementary form in 1821. The general form was discovered by [https://en.wikipedia.org/wiki/Viktor_Bunyakovsky|Viktor Bunyakovsky] in 1849 and independently by [https://en.wikipedia.org/wiki/Hermann_Amandus_Schwarz|Hermann Schwarz] in 1888. | ||
== Elementary Form == | == Elementary Form == | ||
For any real numbers <math> a_1, \ldots, a_n </math> and <math> b_1, \ldots, b_n </math>, | For any real numbers <math> a_1, \ldots, a_n </math> and <math> b_1, \ldots, b_n </math>, | ||
− | < | + | <cmath> |
− | + | \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) | |
− | \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 ) | + | </cmath> |
− | </ | + | with equality when there exists a nonzero constant <math>\mu</math> such that for all <math> 1 \le i \le n </math>, <math>\mu a_i = b_i </math>. |
− | |||
− | with equality when there | ||
− | === | + | === Discussion === |
− | + | Consider the vectors <math> \mathbf{a} = \langle a_1, \ldots a_n \rangle </math> and <math> {} \mathbf{b} = \langle b_1, \ldots b_n \rangle </math>. If <math>\theta </math> is the [[angle]] formed by <math> \mathbf{a} </math> and <math> \mathbf{b} </math>, then the left-hand side of the inequality is equal to the square of the [[dot product]] of <math>\mathbf{a} </math> and <math> \mathbf{b} </math>, or <math>(\mathbf{a} \cdot \mathbf{b})^2 = a^2 b^2 (\cos\theta) ^2</math> .The right hand side of the inequality is equal to <math> \left( ||\mathbf{a}|| * ||\mathbf{b}|| \right)^2 = a^2b^2</math>. The inequality then follows from <math> |\cos\theta | \le 1 </math>, with equality when one of <math> \mathbf{a,b} </math> is a multiple of the other, as desired. | |
− | |||
− | Consider the vectors <math> \mathbf{a} = \langle a_1, \ldots a_n \rangle </math> and <math> {} \mathbf{b} = \langle b_1, \ldots b_n \rangle </math>. If <math>\theta </math> is the [[angle]] formed by <math> \mathbf{a} </math> and <math> \mathbf{b} </math>, then the left-hand side of the inequality is equal to the square of the [[dot product]] of <math> \mathbf{a} </math> and <math> \mathbf{b} </math>, or <math> | ||
=== Complex Form === | === Complex Form === | ||
Line 22: | Line 20: | ||
Let <math> a_1, \ldots, a_n </math> and <math> b_1, \ldots, b_n </math> be [[complex numbers]]. Then | Let <math> a_1, \ldots, a_n </math> and <math> b_1, \ldots, b_n </math> be [[complex numbers]]. Then | ||
− | < | + | <cmath> |
− | + | \left| \sum_{i=1}^na_ib_i \right|^2 \le \left(\sum_{i=1}^{n}|a_i^2| \right) \left( \sum_{i=1}^n |b_i^2| \right) | |
− | \left| \sum_{i=1}^na_ib_i \right|^2 \le \left (\sum_{i=1}^{n}|a_i^2|\right ) \left (\sum_{i=1}^n |b_i^2|\right ) | + | </cmath> |
− | </ | + | This appears to be more powerful, but it follows from |
− | + | <cmath> | |
− | This appears to be more powerful, but it follows | + | \left| \sum_{i=1}^n a_ib_i \right| ^2 \le \left( \sum_{i=1}^n |a_i| \cdot |b_i| \right)^2 \le \left(\sum_{i=1}^n |a_i^2| \right) \left( \sum_{i=1}^n |b_i^2| \right) |
− | < | + | </cmath> |
− | + | ||
− | \left| \sum_{i=1}^n a_ib_i \right| ^2 \le \left( \sum_{i=1}^n |a_i| \cdot |b_i| \right)^2 \le \left(\sum_{i=1}^n |a_i^2|\right)\left( \sum_{i=1}^n |b_i^2|\right ) | + | == Upper Bound on (Σa)(Σb) == |
− | </math>. | + | |
− | </ | + | Let <math>a_1, a_2, \ldots, a_n</math> and <math>b_1, b_2, \ldots, b_n</math> be two sequences of positive real numbers with |
+ | <cmath> | ||
+ | 0 < m \le \frac{a_i}{b_i} \le M | ||
+ | </cmath> | ||
+ | for <math>1 \le i \le n</math>. Then | ||
+ | <cmath> | ||
+ | \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, | ||
+ | </cmath> | ||
+ | with equality if and only if, for some ordering of the pairs <math>(a_i,b_i) \mapsto (a_{\sigma(i)},b_{\sigma(i)})</math>, some <math>0 \le j \le n</math> exists such that <math>a_{\sigma(i)}=mb_{\sigma(i)}</math> for <math>1 \le \sigma(i) \le j</math> and <math>a_{\sigma(i)}=Mb_{\sigma(i)}</math> for <math>j+1 \le \sigma(i) \le n</math>, and | ||
+ | <cmath> | ||
+ | m\sum_{\sigma(i)=1}^{j}b_{\sigma(i)}^2 = M\sum_{\sigma(i)=j+1}^{n}b_{\sigma(i)}^2. | ||
+ | </cmath> | ||
+ | If we restrict that <math>m_1 \le a_i \le M_1</math> and <math>m_2 \le b_i \le M_2</math> for all <math>i</math>, then it's clear that for <math>a_i/b_i</math> to be <math>m=m_1/M_2</math> or <math>M=M_1/m_2</math> for all <math>i</math>, then <math>a_i=m_1 \Longleftrightarrow b_i=M_2</math> and <math>a_i=M_1 \Longleftrightarrow b_i=m_2</math>, so | ||
+ | <cmath> | ||
+ | m\sum_{\sigma(i)=1}^{j}b_{\sigma(i)}^2 = M\sum_{\sigma(i)=j+1}^{n}b_{\sigma(i)}^2 | ||
+ | </cmath> | ||
+ | is equivalent to | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | m(jM_2^2) = M((n-j)m_2^2) &\Longleftrightarrow m_1M_2j = M_1m_2(n-j)\\ | ||
+ | &\Longleftrightarrow j = \left(\frac{M_1m_2}{M_1m_2+m_1M_2}\right) n. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | (When this is not an integer, the maximum occurs when <math>j</math> is either the ceiling or floor of the right-hand side.) In the special case that <math>a_ib_i = k > 0</math> is constant for all <math>i</math>, we have <math>M_1=1/m_2</math> and <math>m_1=1/M_2</math>, so here <math>j</math> must be <math>n/2</math>. | ||
+ | |||
+ | === Proof === | ||
+ | |||
+ | Note that for all <math>i</math>, we have | ||
+ | <cmath> | ||
+ | 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) | ||
+ | </cmath> | ||
+ | or | ||
+ | <cmath> | ||
+ | (M+m)a_ib_i \ge a_i^2+(Mm)b_i^2, | ||
+ | </cmath> | ||
+ | with equality if and only if <math>a_i=mb_i</math> or <math>a_i=Mb_i</math>. Summing up these inequalities over <math>1 \le i \le n</math>, we obtain from AM-GM that | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | (M+m)\sum_{i=1}^{n}a_ib_i &\ge \sum_{i=1}^{n}a_i^2 + (Mm)\sum_{i=1}^{n}b_i^2\\ | ||
+ | &\ge 2\sqrt{Mm \left(\sum_{i=1}^{n}a_i^2 \right) \left(\sum_{i=1}^{n}b_i^2 \right)}, | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | and squaring gives us the desired bound. For equality to occur, we must have <math>a_i=mb_i</math> or <math>a_i=Mb_i</math> for all <math>i</math>. If, without loss of generality, <math>a_i=mb_i</math> for <math>1 \le i \le j</math> and <math>a_i=Mb_i</math> for <math>j+1 \le i \le n</math> for some <math>0 \le j \le n</math>, then for the AM-GM to reach equality we must have (assume <math>M>m</math> since <math>M=m</math> is trivial) | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \sum_{i=1}^{n}a_i^2 &= Mm\sum_{i=1}^{n}b_i^2\\ | ||
+ | m^2\sum_{i=1}^{j}b_i^2 + M^2\sum_{i=j+1}^{n}b_i^2 &= Mm\sum_{i=1}^{j}b_i^2 + Mm\sum_{i=j+1}^{n}b_i^2\\ | ||
+ | (m-M)m\sum_{i=1}^{j}b_i^2 &= (m-M)M\sum_{i=j+1}^{n}b_i^2\\ | ||
+ | m\sum_{i=1}^{j}b_i^2 &= M\sum_{i=j+1}^{n}b_i^2. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
== General Form == | == General Form == | ||
− | Let <math>V </math> be a [[vector space]], and let <math> \langle \cdot, \cdot \rangle : V \times V \ | + | Let <math>V </math> be a [[vector space]], and let <math> \langle \cdot, \cdot \rangle : V \times V \to \mathbb{R} </math> be an [[inner product]]. Then for any <math> \mathbf{a,b} \in V </math>, |
− | < | + | <cmath> |
− | + | \langle \mathbf{a,b} \rangle^2 \le \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle , | |
− | \langle \mathbf{a,b} \rangle^2 \le \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle | + | </cmath> |
− | |||
− | </ | ||
with equality if and only if there exist constants <math>\mu, \lambda </math> not both zero such that <math> \mu\mathbf{a} = \lambda\mathbf{b} </math>. | with equality if and only if there exist constants <math>\mu, \lambda </math> not both zero such that <math> \mu\mathbf{a} = \lambda\mathbf{b} </math>. | ||
Line 47: | Line 93: | ||
Consider the polynomial of <math> t </math> | Consider the polynomial of <math> t </math> | ||
− | < | + | <cmath> |
− | + | \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 . | |
− | \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 | + | </cmath> |
− | |||
− | </ | ||
This must always be greater than or equal to zero, so it must have a non-positive discriminant, i.e., <math> \langle \mathbf{a,b} \rangle^2 </math> must be less than or equal to <math> \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle </math>, with equality when <math> \mathbf{a = 0} </math> or when there exists some scalar <math>-t </math> such that <math> -t\mathbf{a} = \mathbf{b} </math>, as desired. | This must always be greater than or equal to zero, so it must have a non-positive discriminant, i.e., <math> \langle \mathbf{a,b} \rangle^2 </math> must be less than or equal to <math> \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle </math>, with equality when <math> \mathbf{a = 0} </math> or when there exists some scalar <math>-t </math> such that <math> -t\mathbf{a} = \mathbf{b} </math>, as desired. | ||
Line 57: | Line 101: | ||
We consider | We consider | ||
− | < | + | <cmath> |
− | + | \langle \mathbf{a-b, a-b} \rangle = \langle \mathbf{a,a} \rangle + \langle \mathbf{b,b} \rangle - 2 \langle \mathbf{a,b} \rangle . | |
− | \langle \mathbf{a-b, a-b} \rangle = \langle \mathbf{a,a} \rangle + \langle \mathbf{b,b} \rangle - 2 \langle \mathbf{a,b} \rangle | + | </cmath> |
− | |||
− | </ | ||
Since this is always greater than or equal to zero, we have | Since this is always greater than or equal to zero, we have | ||
− | < | + | <cmath> |
− | + | \langle \mathbf{a,b} \rangle \le \frac{1}{2} \langle \mathbf{a,a} \rangle + \frac{1}{2} \langle \mathbf{b,b} \rangle . | |
− | \langle \mathbf{a,b} \rangle \le \frac{1}{2} \langle \mathbf{a,a} \rangle + \frac{1}{2} \langle \mathbf{b,b} \rangle | + | </cmath> |
− | |||
− | </ | ||
Now, if either <math> \mathbf{a} </math> or <math> \mathbf{b} </math> is equal to <math> \mathbf{0} </math>, then <math> \langle \mathbf{a,b} \rangle^2 = \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle = 0 </math>. Otherwise, we may [[normalize]] so that <math> \langle \mathbf {a,a} \rangle = \langle \mathbf{b,b} \rangle = 1 </math>, and we have | Now, if either <math> \mathbf{a} </math> or <math> \mathbf{b} </math> is equal to <math> \mathbf{0} </math>, then <math> \langle \mathbf{a,b} \rangle^2 = \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle = 0 </math>. Otherwise, we may [[normalize]] so that <math> \langle \mathbf {a,a} \rangle = \langle \mathbf{b,b} \rangle = 1 </math>, and we have | ||
− | < | + | <cmath> |
− | + | \langle \mathbf{a,b} \rangle \le 1 = \langle \mathbf{a,a} \rangle^{1/2} \langle \mathbf{b,b} \rangle^{1/2} , | |
− | \langle \mathbf{a,b} \rangle \le 1 = \langle \mathbf{a,a} \rangle^{1/2} \langle \mathbf{b,b} \rangle^{1/2} | + | </cmath> |
− | |||
− | </ | ||
with equality when <math>\mathbf{a} </math> and <math> \mathbf{b} </math> may be scaled to each other, as desired. | with equality when <math>\mathbf{a} </math> and <math> \mathbf{b} </math> may be scaled to each other, as desired. | ||
+ | |||
+ | === Proof 3 === | ||
+ | |||
+ | Consider <math>a-\lambda b</math> for some scalar <math>\lambda</math>. Then: | ||
+ | <math>0\le||a-\lambda b||^2</math> (by the Trivial Inequality) | ||
+ | <math>=\langle a-\lambda b,a-\lambda b\rangle</math> | ||
+ | <math>=\langle a,a\rangle-2\lambda\langle a,b\rangle+\lambda^2\langle y,y\rangle</math> | ||
+ | <math>=||a||^2-2\lambda\langle a,b\rangle+\lambda^2||b||^2</math>. | ||
+ | Now, let <math>\lambda=\frac{\langle a,b\rangle}{||b||^2}</math>. Then, we have: | ||
+ | <math>0\le||a||^2-\frac{\langle a,b\rangle|^2}{||b||^2}</math> | ||
+ | <math>\implies\langle a,b\rangle|^2\le||a||^2||b||^2=\langle a,a\rangle\cdot\langle b,b\rangle</math>. <math>\square</math> | ||
=== Examples === | === 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 <math> f,g : [a,b] \mapsto \mathbb{R} </math>, | 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 <math> f,g : [a,b] \mapsto \mathbb{R} </math>, | ||
− | < | + | <cmath> |
− | + | \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 | |
− | \ | + | </cmath> |
− | </ | ||
− | |||
with equality when there exist constants <math> \mu, \lambda </math> not both equal to zero such that for <math> t \in [a,b] </math>, | with equality when there exist constants <math> \mu, \lambda </math> not both equal to zero such that for <math> t \in [a,b] </math>, | ||
− | < | + | <cmath> |
− | + | \mu \int_a^t f(x)dx = \lambda \int_a^t g(x)dx . | |
− | \mu \int_a^t f(x)dx = \lambda \int_a^t g(x)dx | + | </cmath> |
− | |||
− | </ | ||
==Problems== | ==Problems== | ||
+ | |||
===Introductory=== | ===Introductory=== | ||
+ | |||
*Consider the function <math>f(x)=\frac{(x+k)^2}{x^2+1},x\in (-\infty,\infty)</math>, where <math>k</math> is a positive integer. Show that <math>f(x)\le k^2+1</math>. ([[User:Temperal/The_Problem_Solver's Resource Competition|Source]]) | *Consider the function <math>f(x)=\frac{(x+k)^2}{x^2+1},x\in (-\infty,\infty)</math>, where <math>k</math> is a positive integer. Show that <math>f(x)\le k^2+1</math>. ([[User:Temperal/The_Problem_Solver's Resource Competition|Source]]) | ||
+ | * [http://www.mathlinks.ro/Forum/viewtopic.php?t=78687 (APMO 1991 #3)] Let <math>a_1</math>, <math>a_2</math>, <math>\cdots</math>, <math>a_n</math>, <math>b_1</math>, <math>b_2</math>, <math>\cdots</math>, <math>b_n</math> be positive real numbers such that <math>a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n</math>. Show that | ||
+ | <cmath>\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}</cmath> | ||
+ | |||
===Intermediate=== | ===Intermediate=== | ||
− | *Let <math>ABC </math> be a triangle such that | + | |
− | < | + | *Let <math>ABC</math> be a triangle such that |
− | + | <cmath> | |
− | \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 | + | \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 , |
− | + | </cmath> | |
− | </ | + | where <math>s</math> and <math>r</math> denote its [[semiperimeter]] and [[inradius]], respectively. Prove that triangle <math>ABC </math> is similar to a triangle <math>T </math> whose side lengths are all positive integers with no common divisor and determine those integers. |
− | where <math>s </math> and <math>r </math> denote its [[semiperimeter]] and [[inradius]], respectively. Prove that triangle <math>ABC </math> is similar to a triangle <math>T </math> whose side lengths are all positive integers with no common divisor and determine those integers. | ||
([[2002 USAMO Problems/Problem 2|Source]]) | ([[2002 USAMO Problems/Problem 2|Source]]) | ||
+ | |||
===Olympiad=== | ===Olympiad=== | ||
+ | |||
*<math>P</math> is a point inside a given triangle <math>ABC</math>. <math>D, E, F</math> are the feet of the perpendiculars from <math>P</math> to the lines <math>BC, CA, AB</math>, respectively. Find all <math>P</math> for which | *<math>P</math> is a point inside a given triangle <math>ABC</math>. <math>D, E, F</math> are the feet of the perpendiculars from <math>P</math> to the lines <math>BC, CA, AB</math>, respectively. Find all <math>P</math> for which | ||
− | + | <cmath> | |
− | < | ||
− | |||
\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF} | \frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF} | ||
− | </ | + | </cmath> |
− | |||
− | |||
is least. | is least. | ||
Line 124: | Line 171: | ||
+ | [[Category:Algebra]] | ||
[[Category:Inequality]] | [[Category:Inequality]] | ||
[[Category:Theorems]] | [[Category:Theorems]] |
Latest revision as of 14:57, 17 January 2021
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.
Contents
Elementary Form
For any real numbers and
,
with equality when there exists a nonzero constant
such that for all
,
.
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.
Let and
be complex numbers. Then
This appears to be more powerful, but it follows from
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
.
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)
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
.
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.
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.
Proof 3
Consider for some scalar
. Then:
(by the Trivial Inequality)
.
Now, let
. Then, we have:
.
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
,
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
Intermediate
- Let
be a triangle such that
where
and
denote its semiperimeter and inradius, respectively. Prove that triangle
is similar to a triangle
whose side lengths are all positive integers with no common divisor and determine those integers.
(Source)
Olympiad
is a point inside a given triangle
.
are the feet of the perpendiculars from
to the lines
, respectively. Find all
for which
is least.
(Source)
Other Resources
Books
- The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities by J. Michael Steele.
- Problem Solving Strategies by Arthur Engel contains significant material on inequalities.