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−  ==Introduction==
 
−  SINCE MY COMPUTER WON'T LOAD THIS FOR SOME REASON, I'LL BE UPDATING THIS AS I GO THOUGH :)
 
   
−  Ok, so inspired by master math solver Lcz, I have decided to take Oly notes (for me) online! I'll probably be yelled at even more for staring at the computer, but I know that this is for my good. (Also this thing is almost the exact same format as Lcz's :P ). (Ok, actually, a LOT of credits to Lcz)
 
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−  ==Algebra==
 
−  Problems worth noting/reviewing
 
−  I'll leave this empty for now, I want to start on HARD stuff yeah!
 
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−  ===Inequalities===
 
−  We shall begin with INEQUALITIES! They should be fun enough. I should probably begin with some theorems.
 
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−  ====Power mean (special case)====
 
−  Statement: Given that <math>a_1, a_2, a_3, ... a_n > 0</math>, <math>a_{i} \in \mathbb{R}</math> where <math>1 \le i \le n</math>. Define the <math>pm_x(a_1, a_2, \cdots , a_n)</math> as:<cmath>(\frac{a_1^x+a_2^x+\cdots+a_n^x}{n})^{\frac{1}{x}},</cmath>where <math>x\neq0</math>, and:<cmath>\sqrt[n]{a_{1}a_{2}a_{3} \cdots a_{n}}.</cmath>where <math>x=0</math>.
 
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−  If <math>x \ge y</math>, then<cmath>pm_x(a_1, a_2, \cdots , a_n) \ge pm_y(a_1, a_2, \cdots , a_n).</cmath>
 
−  Power mean (weighted)
 
−  Statement: Let <math>a_1, a_2, a_3, . . . a_n</math> be positive real numbers. Let <math>w_1, w_2, w_3, . . . w_n</math> be positive real numbers ("weights") such that <math>w_1+w_2+w_3+ . . . w_n=1</math>. For any <math>r \in \mathbb{R}</math>,
 
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−  if <math>r=0</math>,
 
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−  <cmath>P(r)=a_1^{w_1} a_2^{w_2} a_3^{w_3} . . . a_n^{w_n}</cmath>.
 
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−  if <math>r \neq 0</math>,
 
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−  <cmath>P(r)=(w_1a_1^r+w_2a_2^r+w_3a_3^r . . . +w_na_n^r)^{\frac{1}{r}}</cmath>.
 
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−  If <math>r>s</math>, then <math>P(r) \geq P(s)</math>. Equality occurs if and only if all the <math>a_i</math> are equal.
 
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−  ====CauchySwartz Inequality====
 
−  Let there be two sets of integers, <math>a_1, a_2, \cdots a_n</math> and <math>b_1, b_2, \cdots b_n</math>, such that <math>n</math> is a positive integer, where all members of the sequences are real, then we have:<cmath>(a_1^2+a_2^2+\cdots +a_n^2)(b_1^2+b_2^2+ \cdots +b_n^2)\ge (a_1b_1 + a_2b_2 + \cdots +a_nb_n)^2.</cmath>Equality holds if for all <math>a_i</math>, where <math>1\le i \le n</math>, <math>a_i=0</math>, or for all <math>b_i</math>, where <math>1\le i \le n</math>, <math>b_i=0</math>., or we have some constant <math>k</math> such that <math>b_i=ka_i</math> for all <math>i</math>.
 
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−  ====Bernoulli's Inequality====
 
−  Given that <math>n</math>, <math>x</math> are real numbers such that <math>n\ge 0</math> and <math>x \ge 1</math>, we have:<cmath>(1+x)^n \ge 1+nx.</cmath>
 
−  Rearrangement Inequality
 
−  Given that<cmath>x_1 \ge x_2 \ge x_3 \cdots x_n</cmath>and<cmath>y_1 \ge y_2 \ge y_3 \cdots y_n.</cmath>We have:<cmath>x_1y_1+x_2y_2 + \cdots + x_ny_n</cmath>is greater than any other pairings' sum.
 
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−  ====Holder's Inequality====
 
−  If <math>a_1, a_2, \cdots, a_n</math>, <math>b_1, b_2, \cdots, b_n</math>, <math>\cdots</math>, <math>z_1, z_2, \cdots, z_n</math> are nonnegative real numbers and <math>\lambda_a, \lambda_b, \cdots, \lambda_z</math> are nonnegative reals with sum of <math>1</math>, then:<cmath>a_1^{\lambda_a}b_1^{\lambda_b} \cdots z_1^{\lambda_z} + \cdots + a_n^{\lambda_a} b_n^{\lambda_b} \cdots z_n^{\lambda_z} \le (a_1 + \cdots + a_n)^{\lambda_a} (b_1 + \cdots + b_n)^{\lambda_b} \cdots (z_1 + \cdots + z_n)^{\lambda_z} .</cmath>This is a generalization of the Cauchy Swartz Inequality.
 
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−  ==Combinatorics==
 
−  ==Number Theory==
 
−  ==Geometry==
 