# Difference between revisions of "User:Lcz"

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− | + | ==Introduction== | |

+ | So I failed the AOIME. I don't really want to do any more stupid AIME prep, so I have decided to go do some Oly prep :) | ||

+ | ==Inequalities== | ||

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

− | + | [b]Basics[/b] | |

+ | 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 <math>a,b,c>0</math> with <math>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1</math>. Prove that <math>(a+1)(b+1)(c+1) \geq 64</math> | ||

+ | |||

+ | Solution: | ||

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

+ | |||

+ | <math>\frac{a+1}{a}+\frac{b+1}{b}+\frac{c+1}{c}=4</math>, <math>\frac{(a+1)(b+1)(c+1)}{abc} \geq \frac{64}{27}</math>, | ||

+ | <math>abc \geq 27</math> | ||

+ | |||

+ | By AM-GM. | ||

+ | Obviously the trivial solution <math>(3,3,3)</math> 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 | ||

+ | <math>(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^3=1</math>, we get | ||

+ | |||

+ | <math>\frac{(ab+bc+ac)^3}{(abc)^2} \geq 27</math>, <math>(\frac{ab+bc+ac}{3})^3 \geq (abc)^2</math>, <math>(\frac{ab+bc+ac}{3}) \geq (abc)^{2/3}</math> | ||

+ | |||

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

+ | |||

+ | Example 2: | ||

+ | |||

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

+ | |||

+ | Solution: | ||

+ | First we homogenize: | ||

+ | |||

+ | <math>a^2+b^2+c^2 \geq (a+b+c)(abc)^{1/3}</math> | ||

+ | |||

+ | Which is true because <math>(2,0,0)</math> majorizes <math>(\frac{4}{3}, \frac{1}{3}, \frac{1}{3})</math> | ||

+ | |||

+ | |||

+ | Cauchy: | ||

+ | |||

+ | Problem 1: (2009 usamo/4): For <math>n \ge 2</math> let <math>a_1</math>, <math>a_2</math>, ..., <math>a_n</math> be positive real numbers such that | ||

+ | <math> (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 </math> | ||

+ | Prove that <math>\text{max}(a_1, a_2, ... ,a_n) \le 4 \text{min}(a_1, a_2, ... , a_n)</math>. | ||

+ | |||

+ | Try to solve this on your own! Very cute problem. | ||

+ | Note that you'll probably only ever need Holder's for <math>3</math> variables... | ||

+ | |||

+ | Example 3 (2004 usamo/5) | ||

+ | |||

+ | Let <math>a</math>, <math>b</math>, and <math>c</math> be positive real numbers. Prove that | ||

+ | <math>(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \ge (a+b+c)^3</math>. | ||

+ | |||

+ | Solution: | ||

+ | |||

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

+ | |||

+ | 2. Now all we have to prove is that <math>a^5-a^3-a^2+1 \geq 0</math>, or <math>(a^3-1)(a^2-1) \geq 0</math>. | ||

+ | Now note that if <math>a<1</math>, this is true, if <math>a=1</math>, this is true, and if <math>a>1</math>, this is true as well, and as we have exhausted all cases, we are done. | ||

+ | |||

+ | |||

+ | [b]More advanced stuff, learn some calculus[/b] | ||

+ | |||

+ | You will need to know derivatives for this part. It's actually pretty simple. | ||

+ | |||

+ | Basically, the derivative of <math>x^n</math> is <math>nx^{n-1}</math> | ||

+ | |||

+ | For the rest of this to make sense, you also need to know what it is. The derivative is what a function's "slope" is, as in, if you take all the pairs of two points it would be the best "curved" line that would best match the slopes combined. Oops I can't really explain this, but you probably don't get it. | ||

+ | |||

+ | 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. | ||

+ | |||

+ | |||

+ | Jensen's inequality says that if <math>f(x)</math> is a convex function in the interval <math>I</math>, for all <math>a_i</math> in <math>I</math>, | ||

+ | <math>\frac{ f(a_1)+f(a_2)+f(a_3)...f(a_n)}{n} \geq f(\frac{a_1+a_2+a_3. . .}{n})</math> | ||

+ | |||

+ | |||

+ | Karamata's inequality says that if <math>f(x)</math> is convex in the interval <math>I</math>, the sequence <math>(x_n)</math> majorizes <math>(y_n)</math>,, and all <math>x_i, y_i</math> are in <math>I</math>, | ||

+ | <math>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)</math> | ||

+ | |||

+ | |||

+ | Problem 2: Using Jensen's and Holder's, solve 2001 IMO/2: | ||

+ | Let <math>a,b,c</math> be positive real numbers. Prove <math>\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1</math>. | ||

+ | |||

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

+ | |||

+ | [b] Function Equations [/b] |

## Revision as of 13:36, 9 June 2020

## Introduction

So I failed the AOIME. I don't really want to do any more stupid AIME prep, so I have decided to go do some Oly prep :)

## Inequalities

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

[b]Basics[/b]

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 with . Prove that

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

, ,

By AM-GM. Obviously the trivial solution 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 , we get

, ,

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

Example 2:

Let (Again Evan Chen) and . Prove that .

Solution: First we homogenize:

Which is true because majorizes

Cauchy:

Problem 1: (2009 usamo/4): For let , , ..., be positive real numbers such that Prove that .

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

Example 3 (2004 usamo/5)

Let , , and be positive real numbers. Prove that

.

Solution:

1. The is cubed, so we try to use Holder's. The simplest way to do this is just to use on the LHS.

2. Now all we have to prove is that , or . Now note that if , this is true, if , this is true, and if , this is true as well, and as we have exhausted all cases, we are done.

[b]More advanced stuff, learn some calculus[/b]

You will need to know derivatives for this part. It's actually pretty simple.

Basically, the derivative of is

For the rest of this to make sense, you also need to know what it is. The derivative is what a function's "slope" is, as in, if you take all the pairs of two points it would be the best "curved" line that would best match the slopes combined. Oops I can't really explain this, but you probably don't get it.

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.

Jensen's inequality says that if is a convex function in the interval , for all in ,

Karamata's inequality says that if is convex in the interval , the sequence majorizes ,, and all are in ,

Problem 2: Using Jensen's and Holder's, solve 2001 IMO/2:
Let be positive real numbers. Prove .

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

[b] Function Equations [/b]