Difference between revisions of "User:Temperal/The Problem Solver's Resource11"
m (→Holder's Inequality: typo?) |
(→Trivial Inequality: proof) |
||
(6 intermediate revisions by the same user not shown) | |||
Line 1: | Line 1: | ||
__NOTOC__ | __NOTOC__ | ||
− | + | {{User:Temperal/testtemplate|page 11}} | |
− | + | ==<span style="font-size:20px; color: blue;">Inequalities</span>== | |
− | + | My favorite topic, saved for last. | |
− | + | ===Trivial Inequality=== | |
− | + | For any real <math>x</math>, <math>x^2\ge 0</math>, with equality iff <math>x=0</math>. | |
− | ==< | + | |
− | + | Proof: We proceed by contradiction. Suppose there exists a real <math>x</math> such that <math>x^2<0</math>. We can have either <math>x=0</math>, <math>x>0</math>, or <math>x<0</math>. If <math>x=0</math>, then there is a clear contradiction, as <math>x^2 = 0^2 \not < 0</math>. If <math>x>0</math>, then <math>x^2 < 0</math> gives <math>x < \frac{0}{x} = 0</math> upon division by <math>x</math> (which is positive), so this case also leads to a contradiction. Finally, if <math>x<0</math>, then <math>x^2 < 0</math> gives <math>x > \frac{0}{x} = 0</math> upon division by <math>x</math> (which is negative), and yet again we have a contradiction. | |
+ | |||
+ | Therefore, <math>x^2 \ge 0</math> for all real <math>x</math>, as claimed. | ||
+ | |||
+ | ===Arithmetic Mean/Geometric Mean Inequality=== | ||
+ | For any set of real numbers <math>S</math>, <math>\frac{S_1+S_2+S_3....+S_{k-1}+S_k}{k}\ge \sqrt[k]{S_1\cdot S_2 \cdot S_3....\cdot S_{k-1}\cdot S_k}</math> with equality iff <math>S_1=S_2=S_3...=S_{k-1}=S_k</math>. | ||
+ | |||
+ | |||
+ | ===Cauchy-Schwarz Inequality=== | ||
+ | |||
+ | For any real numbers <math>a_1,a_2,...,a_n</math> and <math>b_1,b_2,...,b_n</math>, the following holds: | ||
+ | |||
+ | <math>\left(\sum a_i^2\right)\left(\sum b_i^2\right) \ge \left(\sum a_ib_i\right)^2</math> | ||
+ | |||
+ | ====Cauchy-Schwarz Variation==== | ||
+ | |||
+ | For any real numbers <math>a_1,a_2,...,a_n</math> and positive real numbers <math>b_1,b_2,...,b_n</math>, the following holds: | ||
+ | |||
+ | <math>\sum\left({{a_i^2}\over{b_i}}\right) \ge {{\sum a_i^2}\over{\sum b_i}}</math>. | ||
+ | |||
+ | ===Power Mean Inequality=== | ||
+ | |||
+ | Take a set of functions <math>m_j(a) = \left({\frac{\sum a_i^j}{n}}\right)^{1/j}</math>. | ||
+ | |||
+ | Note that <math>m_0</math> does not exist. The geometric mean is <math>m_0 = \lim_{k \to 0} m_k</math>. | ||
+ | For non-negative real numbers <math>a_1,a_2,\ldots,a_n</math>, the following holds: | ||
+ | |||
+ | <math>m_x \le m_y</math> for reals <math>x<y</math>. | ||
+ | |||
+ | , if <math>m_2</math> is the quadratic mean, <math>m_1</math> is the arithmetic mean, <math>m_0</math> the geometric mean, and <math>m_{-1}</math> the harmonic mean. | ||
+ | |||
+ | ===RSM-AM-GM-HM Inequality=== | ||
+ | For any positive real numbers <math>x_1,\ldots,x_n</math>: | ||
+ | |||
+ | <math>\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n}\ge\sqrt[n]{x_1\cdots x_n}\ge\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}</math> | ||
+ | |||
+ | with equality iff <math>x_1=x_2=\cdots=x_n</math>. | ||
+ | |||
+ | ===Chebyshev's Inequality=== | ||
+ | |||
+ | Given real numbers <math>a_1 \ge a_2 \ge ... \ge a_n \ge 0</math> and <math>b_1 \ge b_2 \ge ... \ge b_n</math>, we have | ||
+ | |||
+ | <math>{\frac{\sum a_ib_i}{n}} \ge {\frac{\sum a_i}{n}}{\frac{\sum b_i}{n}}</math>. | ||
+ | |||
+ | ===Minkowski's Inequality=== | ||
+ | |||
+ | Given real numbers <math>a_1,a_2,...,a_n</math> and <math>b_1,b_2,\ldots,b_n</math>, the following holds: | ||
+ | |||
+ | <math>\sqrt{\sum a_i^2} + \sqrt{\sum b_i^2} \ge \sqrt{\sum (a_i+b_i)^2}</math> | ||
+ | |||
+ | ===Nesbitt's Inequality=== | ||
+ | |||
+ | For all positive real numbers <math>a</math>, <math>b</math> and <math>c</math>, the following holds: | ||
+ | |||
+ | <math>{\frac{a}{b+c}} + {\frac{b}{c+a}} + {\frac{c}{a+b}} \ge {\frac{3}{2}}</math>. | ||
+ | |||
+ | ===Schur's inequality=== | ||
+ | |||
+ | Given positive real numbers <math>a,b,c</math> and real <math>r</math>, the following holds: | ||
+ | |||
+ | <math>a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b)\ge 0</math>. | ||
+ | |||
===Jensen's Inequality=== | ===Jensen's Inequality=== | ||
For a convex function <math>f(x)</math> and real numbers <math>a_1,a_2,a_3,a_4\ldots,a_n</math> and <math>x_1,x_2,x_3,x_4\ldots,x_n</math>, the following holds: | For a convex function <math>f(x)</math> and real numbers <math>a_1,a_2,a_3,a_4\ldots,a_n</math> and <math>x_1,x_2,x_3,x_4\ldots,x_n</math>, the following holds: | ||
Line 29: | Line 90: | ||
with equality exactly iff all <math>x_i</math> are equivalent. | with equality exactly iff all <math>x_i</math> are equivalent. | ||
− | === | + | ===MacLaurin's Inequality=== |
For non-negative real numbers <math>x_1,x_2,x_3 \ldots, x_n</math>, and <math>d_1,d_2,d_3 \ldots, d_n</math> such that | For non-negative real numbers <math>x_1,x_2,x_3 \ldots, x_n</math>, and <math>d_1,d_2,d_3 \ldots, d_n</math> such that | ||
− | <cmath>d_k = \frac{\ | + | <cmath>d_k = \frac{\sum\limits_{ 1\leq i_1 < i_2 < \cdots < i_k \leq n}x_{i_1} x_{i_2} \cdots x_{i_k}}{{n \choose k}}</cmath>, for <math>k\subset [1,n]</math> the following holds: |
<cmath>d_1 \ge \sqrt[2]{d_2} \ge \sqrt[3]{d_3}\ldots \ge \sqrt[n]{d_n}</cmath> | <cmath>d_1 \ge \sqrt[2]{d_2} \ge \sqrt[3]{d_3}\ldots \ge \sqrt[n]{d_n}</cmath> | ||
Line 39: | Line 100: | ||
[[User:Temperal/The Problem Solver's Resource10|Back to page 10]] | Last page (But also see the | [[User:Temperal/The Problem Solver's Resource10|Back to page 10]] | Last page (But also see the | ||
[[User:Temperal/The Problem Solver's Resource Tips and Tricks|tips and tricks page]], and the | [[User:Temperal/The Problem Solver's Resource Tips and Tricks|tips and tricks page]], and the | ||
− | [[User:Temperal/The Problem Solver's Resource | + | [[User:Temperal/The Problem Solver's Resource Proofs|methods of proof]]! |
− |
Latest revision as of 22:01, 10 January 2009
Introduction | Other Tips and Tricks | Methods of Proof | You are currently viewing page 11. |
Inequalities
My favorite topic, saved for last.
Trivial Inequality
For any real , , with equality iff .
Proof: We proceed by contradiction. Suppose there exists a real such that . We can have either , , or . If , then there is a clear contradiction, as . If , then gives upon division by (which is positive), so this case also leads to a contradiction. Finally, if , then gives upon division by (which is negative), and yet again we have a contradiction.
Therefore, for all real , as claimed.
Arithmetic Mean/Geometric Mean Inequality
For any set of real numbers , with equality iff .
Cauchy-Schwarz Inequality
For any real numbers and , the following holds:
Cauchy-Schwarz Variation
For any real numbers and positive real numbers , the following holds:
.
Power Mean Inequality
Take a set of functions .
Note that does not exist. The geometric mean is . For non-negative real numbers , the following holds:
for reals .
, if is the quadratic mean, is the arithmetic mean, the geometric mean, and the harmonic mean.
RSM-AM-GM-HM Inequality
For any positive real numbers :
with equality iff .
Chebyshev's Inequality
Given real numbers and , we have
.
Minkowski's Inequality
Given real numbers and , the following holds:
Nesbitt's Inequality
For all positive real numbers , and , the following holds:
.
Schur's inequality
Given positive real numbers and real , the following holds:
.
Jensen's Inequality
For a convex function and real numbers and , the following holds:
Holder's Inequality
For positive real numbers , the following holds:
Muirhead's Inequality
For a sequence that majorizes a sequence , then given a set of positive integers , the following holds:
Rearrangement Inequality
For any multi sets and , is maximized when is greater than or equal to exactly of the other members of , then is also greater than or equal to exactly of the other members of .
Newton's Inequality
For non-negative real numbers and the following holds:
,
with equality exactly iff all are equivalent.
MacLaurin's Inequality
For non-negative real numbers , and such that , for the following holds:
with equality iff all are equivalent.
Back to page 10 | Last page (But also see the tips and tricks page, and the methods of proof!