## 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)

## Algebra

Problems worth noting/reviewing I'll leave this empty for now, I want to start on HARD stuff yeah!

### Inequalities

We shall begin with INEQUALITIES! They should be fun enough. I should probably begin with some theorems.

#### Power mean (special case)

Statement: Given that $a_1, a_2, a_3, ... a_n > 0$, $a_{i} \in \mathbb{R}$ where $1 \le i \le n$. Define the $pm_x(a_1, a_2, \cdots , a_n)$ as: $$(\frac{a_1^x+a_2^x+\cdots+a_n^x}{n})^{\frac{1}{x}},$$where $x\neq0$, and: $$\sqrt[n]{a_{1}a_{2}a_{3} \cdots a_{n}}.$$where $x=0$.

If $x \ge y$, then $$pm_x(a_1, a_2, \cdots , a_n) \ge pm_y(a_1, a_2, \cdots , a_n).$$ Power mean (weighted) Statement: Let $a_1, a_2, a_3, . . . a_n$ be positive real numbers. Let $w_1, w_2, w_3, . . . w_n$ be positive real numbers ("weights") such that $w_1+w_2+w_3+ . . . w_n=1$. For any $r \in \mathbb{R}$,

if $r=0$, $$P(r)=a_1^{w_1} a_2^{w_2} a_3^{w_3} . . . a_n^{w_n}$$.

if $r \neq 0$, $$P(r)=(w_1a_1^r+w_2a_2^r+w_3a_3^r . . . +w_na_n^r)^{\frac{1}{r}}$$.

If $r>s$, then $P(r) \geq P(s)$. Equality occurs if and only if all the $a_i$ are equal.

#### Cauchy-Swartz Inequality

Let there be two sets of integers, $a_1, a_2, \cdots a_n$ and $b_1, b_2, \cdots b_n$, such that $n$ is a positive integer, where all members of the sequences are real, then we have: $$(a_1^2+a_2^2+\cdots +a_n^2)(b_1^2+b_2^2+ \cdots +b_n^2)\ge (a_1b_1 + a_2b_2 + \cdots +a_nb_n)^2.$$Equality holds if for all $a_i$, where $1\le i \le n$, $a_i=0$, or for all $b_i$, where $1\le i \le n$, $b_i=0$., or we have some constant $k$ such that $b_i=ka_i$ for all $i$.

#### Bernoulli's Inequality

Given that $n$, $x$ are real numbers such that $n\ge 0$ and $x \ge -1$, we have: $$(1+x)^n \ge 1+nx.$$ Rearrangement Inequality Given that $$x_1 \ge x_2 \ge x_3 \cdots x_n$$and $$y_1 \ge y_2 \ge y_3 \cdots y_n.$$We have: $$x_1y_1+x_2y_2 + \cdots + x_ny_n$$is greater than any other pairings' sum.

#### Holder's Inequality

If $a_1, a_2, \cdots, a_n$, $b_1, b_2, \cdots, b_n$, $\cdots$, $z_1, z_2, \cdots, z_n$ are nonnegative real numbers and $\lambda_a, \lambda_b, \cdots, \lambda_z$ are nonnegative reals with sum of $1$, then: $$a_1^{\lambda_a}b_1^{\lambda_b} \cdots z_1^{\lambda_z} + \cdots + a_n^{\lambda_a} b_n^{\lambda_b} \cdots z_n^{\lambda_z} \le (a_1 + \cdots + a_n)^{\lambda_a} (b_1 + \cdots + b_n)^{\lambda_b} \cdots (z_1 + \cdots + z_n)^{\lambda_z} .$$This is a generalization of the Cauchy Swartz Inequality.