User:Lcz

Introduction

So I failed the AOIME. I don't really want to do any more AIME prep, so I have decided to go do some Oly prep :). Here's some oly notes/favorite problems

Disclaimer

Pretty much all of these problems are from the OTIS excerpts. I try to make mines easier to understand though you should probably just read the OTIS excerpts. These notes are to help me learn, though I don't mind if you read them.

Inequalities

Yay! I love inequalities. Clever algebraic manipulation+thereoms is all you need. It all comes from experience though...

I.Basics

AM-GM, Cauchy (Titu's Lemma as well), Muirhead, and Holder's. These are cool, remember that these should only be used when the inequality is homogenized already.. These are all pretty easy to prove as well.

Example 1: (Evan Chen) Let $a,b,c>0$ with $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ . Prove that $(a+1)(b+1)(c+1) \geq 64$

Solution: We need to try to homogenize this somehow. Plugging in the expression for on the LHS for $1$ won't work. If we try to do something on the left side, we'll still have a degree $3>-1$ . Wait a second, why are they all $a+1$ 's? Let's try to get rid of the $a+1$ 's first. Well, if we add $3$ to both sides of the given condition, we get

$\frac{a+1}{a}+\frac{b+1}{b}+\frac{c+1}{c}=4$ , $\frac{(a+1)(b+1)(c+1)}{abc} \geq \frac{64}{27}$ , $abc \geq 27$

By AM-GM. Obviously the trivial solution $(3,3,3)$ satisfies this, so we haven't made any silly mistakes. We still haven't homogenized, but now the path is clear. Multiplying both sides by $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^3=1$ , we get

$\frac{(ab+bc+ac)^3}{(abc)^2} \geq 27$ , $(\frac{ab+bc+ac}{3})^3 \geq (abc)^2$ , $(\frac{ab+bc+ac}{3}) \geq (abc)^{2/3}$

Which is true from AM-GM. We shall now introduce Muirhead's...

Example 2:

Let $a,b,c>0$ (Again Evan Chen) and $abc=1$ . Prove that $a^2+b^2+c^2 \geq a+b+c$ .

Solution: First we homogenize:

$a^2+b^2+c^2 \geq (a+b+c)(abc)^{1/3}$

Which is true because $(2,0,0)$ majorizes $(\frac{4}{3}, \frac{1}{3}, \frac{1}{3})$

Cauchy:

Problem 1: (2009 usamo/4): For $n \ge 2$ let $a_1$ , $a_2$ , ..., $a_n$ be positive real numbers such that $(a_1+a_2+ ... +a_n)\left( {1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n} \right) \le \left(n+ {1 \over 2} \right) ^2$ Prove that $\text{max}(a_1, a_2, ... ,a_n) \le 4 \text{min}(a_1, a_2, ... , a_n)$ .

Try to solve this on your own! Very cute problem. Note that you'll probably only ever need Holder's for $3$ variables...

Example 3 (2004 usamo/5)

Let $a$ , $b$ , and $c$ be positive real numbers. Prove that $(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \ge (a+b+c)^3$ .

Solution:

1. The $(a+b+c)$ is cubed, so we try to use Holder's. The simplest way to do this is just to use $(a^3+1+1) . . .$ on the LHS.

2. Now all we have to prove is that $a^5-a^3-a^2+1 \geq 0$ , or $(a^3-1)(a^2-1) \geq 0$ . Now note that if $a<1$ , this is true, if $a=1$ , this is true, and if $a>1$ , this is true as well, and as we have exhausted all cases, we are done.

I.More advanced stuff, learn some calculus

You will need to know derivatives for this part. It's actually quite simple. Derivative=Tangent.

Basically, the derivative of $x^n$ is $nx^{n-1}$ Adding and other stuff works in the same way. You also probably need to know

-Product Rule: $(fg)'=f'g+fg'$

-Quotient Rule: $(f/g)'= \frac{g'f-gf'}{g^2}$

-Summing: $(f+g)'=f'+g'$

The second derivative of a function is just applying the derivative twice. A function is convex on an interval if it's second derivative is always positive in that interval. A function is convex if it's second derivative is negative in that interval.

Jensen's inequality says that if $f(x)$ is a convex function in the interval $I$ , for all $a_i$ in $I$ , $\frac{ f(a_1)+f(a_2)+f(a_3)...f(a_n)}{n} \geq f(\frac{a_1+a_2+a_3. . .a_n}{n})$ The opposite holds if $f(x)$ is concave.

Karamata's inequality says that if $f(x)$ is convex in the interval $I$ , the sequence $(x_n)$ majorizes $(y_n)$ ,, and all $x_i, y_i$ are in $I$ , $f(x_1)+f(x_2)+f(x_3) . . . f(x_n) \geq f(y_1)+f(y_2)+f(y_3). . . f(y_n)$

TLT (Tangent Line Trick) is basically where you either

a. take the derivative, and plug in the equality cases or

b. plugging in both equality cases to form a line.

Problem 2: Show that $\frac{1}{\sqrt5}+\frac{1}{\sqrt4}+\frac{1}{\sqrt2}>\frac{1}{\sqrt4}+\frac{1}{\sqrt4}+\frac{1}{\sqrt3}$

Problem 3: Using Jensen's and Holder's, solve 2001 IMO/2: Let $a,b,c$ be positive real numbers. Prove $\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1$ .

Problem 4: (2017 usamo/6) Find the minimum possible value of $\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4},$ given that $a,b,c,d,$ are nonnegative real numbers such that $a+b+c+d=4$ .

Problem 5: (Japanese MO 1997/6) Prove that

$\frac{\left(b+c-a\right)^{2}}{\left(b+c\right)^{2}+a^{2}}+\frac{\left(c+a-b\right)^{2}}{\left(c+a\right)^{2}+b^{2}}+\frac{\left(a+b-c\right)^{2}}{\left(a+b\right)^{2}+c^{2}}\geq\frac35$

for any positive real numbers $a$ , $b$ , $c$ .

I.Extra

Ravi Substitution for triangles. $(a,b,c)=(x+y, y+z, x+z)$ .

Weighted AM-GM (This is kind of dumb imo, just use AM-GM lol) The weighted form of AM-GM is given by using weighted averages. For example, the weighted arithmetic mean of $x$ and $y$ with $3:1$ is $\frac{3x+1y}{3+1}$ and the geometric is $\sqrt[3+1]{x^3y}$ . (AoPS Wiki)

Problem 6: (1983 imo/6, using Ravi!)

Let $a$ , $b$ and $c$ be the lengths of the sides of a triangle. Prove that

$a^2 b(a-b) + b^2 c(b-c) + c^2 (c-a) \geq 0$ .

Determine when equality occurs.

Inequality Problems:

https://artofproblemsolving.com/wiki/index.php/Category:Olympiad_Inequality_Problems

https://artofproblemsolving.com/wiki/index.php/Category:Olympiad_Algebra_Problems

Function Equations

Oops...I kind of suck at these :P

~Lcz 6/9/2020 at 12:49 CST

-Function=anything.

-Symmetry

-Plug in $0$

-Check for linear/constant solutions first. They are usually the only ones.

-fff trick is pro

-Pointwise trap?!?

-Be aware of your domain/range. ( $N, Q, Z, C$ (Probably not))

F.Intro

injective: a-1, b-3, c-2, nothing-4

surjective: a-1, b-1, b-2, c-3

bijective: a-1, b-3, c-2

A function $f(x)$ where its domain and range are same is an involution if $f(f(x)) = x$ for every $x$ in its range/domain. This function is bijective.

F.Ok these things are sort of cool

Example 1 (2002 usamo/4):

Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that

$f(x^2 - y^2) = xf(x) - yf(y)$

for all pairs of real numbers $x$ and $y$ .

Solution:

I claim that the sole solution is $f(x)=cx$ for some constant $c$ .

First note that setting $x$ and $y$ as zero respectively yields $f(x^2)=xf(x)$ and $f(-y^2)=-yf(y)$ . Letting $y=x$ in the second equation means that $f$ is odd.

Now, we can substitute into the original equation our findings:

$f(x^2-y^2)=f(x^2)-f(y^2) \implies f(x^2-y^2)+f(y^2)=f(x^2) \implies f(a)+f(b)=f(a+b)$ where $a=x^2-y^2, b=y^2$ .

Finally,

$f(x^2+2x+1)=(x+1)(f(x+1))$

$\implies f(x^2)+f(2x)+f(1)=(x+1)(f(1))+(x+1)(f(x))$

$\implies xf(x)+2f(x)+f(1)=(x+1)(f(1))+(x+1)(f(x))$

$\implies f(x)=xf(1)$ . Setting $f(1)=c$ , we get the desired. $\blacksquare$

F.Actually they kind of suck

Problem 1 (2009 imo/5):

Determine all functions $f$ from the set of positive integers to the set of positive integers such that, for all positive integers $a$ and $b$ , there exists a non-degenerate triangle with sides of lengths

$a,f(b)$ and $f(b+f(a)-1)$ .

This question...is...cool...?

Example 2 (ISL/2015): Determine all functions

$f:\mathbb{Z}\rightarrow\mathbb{Z}$

with the property that

$f(x-f(y))=f(f(x))-f(y)-1$

holds for all $x,y\in\mathbb{Z}$ .

BOGUS Solution:

The answer is $f(x)=-1$ or $f(x)=x+1$ .

Claim: there exists a $y$ such that $f(y)=-1$ .

Proof: set $(x,y)=(x, f(x))$ .

Then $f(x+1)=f(f(x))$ by where $f(y)=-1$ . Now $f(x)=c$ for some constant $c$ , or $f(x)=x+1$ .

The former yields $c=-1 \implies \boxed{f(x)=-1}$ , the latter works and yields $\boxed{f(x)=x+1}$ .

Why this is wrong: You can't just assume because $f(x)=f(y), x=y$ or $f(x)=c$ . Remember, a function is anything. Maybe this function is surjective.

Solution (after that step):

Setting $f(f(x))$ as $f(x+1)$ in the given, we get

$f(x-f(y))=f(x+1)-f(y)-1$ . Setting $x=z-1$ gives

$f(z-f(y)-1)=f(z)-f(y)-1$ , or setting $f(y)+1=w$ ,

$f(z+w)=f(z)+w$ . Now look at all the possible values of $w$ . $0$ is in there! Therefore all possible $w$ are of the form $n\mathbb{Z}$ for some $n$ .