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−  ==Introduction==
 +  Currently: 
−  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
 
   
−  ==Inequalities==
 +  Doing OTIS Excerpts; [[Lcz's Oly Notes]] 
−  Yay! I love inequalities. Clever algebraic manipulation+thereoms is all you need. It all comes from experience though...
 
   
−  ==I.Basics==
 +  OTIS Application: Finished! 
   
−  AMGM, Cauchy (Titu's Lemma as well), Muirhead, and Holder's.
 +  Making mock AMC10 :) Coming in August 2020? 
−  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.
 
   
 +  (Only including AMC/AIME/MathCounts things of course): 
   
−  Example 1:
 +  2018 AMC 8: 15? 
−  (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:
 +  2019 AMC 8: 16 :D 
−  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>,
 +  2019 AMC 10A: 88.5 welp 
−  <math>abc \geq 27</math>
 
   
−  By AMGM.
 +  2020 AMC 10A: 108 (4 sillies) 
−  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>
 +  2020 AMC 10B: 111 (4 sillies again welp) 
   
−  Which is true from AMGM. We shall now introduce Muirhead's...
 +  2020 Austin Math Circle Practice Mathcounts (AMCPM): 41 (2nd written), 1st cdr :P 
   
−  Example 2:
 +  2020 AIME I: 8 (3 sillies rip) 
   
−  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>.
 +  2020 Online mc states: 41 (2 sillies lets gooooooo) 
   
−  Solution:
 +  2020 AOIME: We don't talk about this... (i can edit :P ) 
−  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^5a^3a^2+1 \geq 0</math>, or <math>(a^31)(a^21) \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.
 
−   
−   
−  ==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 <math>x^n</math> is <math>nx^{n1}</math>
 
−  Adding and other stuff works in the same way.
 
−  You also probably need to know
 
−   
−  Product Rule: <math>(fg)'=f'g+fg'</math>
 
−   
−  Quotient Rule: <math>(f/g)'= \frac{g'fgf'}{g^2}</math>
 
−   
−  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.
 
−   
−  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. . .a_n}{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>
 
−   
−   
−  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 <math>\frac{1}{\sqrt5}+\frac{1}{\sqrt4}+\frac{1}{\sqrt2}>\frac{1}{\sqrt4}+\frac{1}{\sqrt4}+\frac{1}{\sqrt3}</math>
 
−   
−  Problem 3: 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>.
 
−   
−  Problem 4: (2017 usamo/6)
 
−  Find the minimum possible value of
 
−  <cmath>\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4},</cmath>
 
−  given that <math>a,b,c,d,</math> are nonnegative real numbers such that <math>a+b+c+d=4</math>.
 
−   
−  Problem 5: (Japanese MO 1997/6)
 
−  Prove that
 
−   
−  <math> \frac{\left(b+ca\right)^{2}}{\left(b+c\right)^{2}+a^{2}}+\frac{\left(c+ab\right)^{2}}{\left(c+a\right)^{2}+b^{2}}+\frac{\left(a+bc\right)^{2}}{\left(a+b\right)^{2}+c^{2}}\geq\frac35</math>
 
−   
−  for any positive real numbers <math> a</math>, <math> b</math>, <math> c</math>.
 
−   
−  ==I.Extra==
 
−   
−  Ravi Substitution for triangles. <math>(a,b,c)=(x+y, y+z, x+z)</math>.
 
−   
−  Power Mean:
 
−  The weighted form of AMGM is given by using weighted averages. For example, the weighted arithmetic mean of <math>x</math> and <math>y</math> with <math>3:1</math> is <math>\frac{3x+1y}{3+1}</math> and the geometric is <math>\sqrt[3+1]{x^3y}</math>. (AoPS Wiki)
 
−   
−  More practice here:
 
−  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 <math>0</math>
 
−   
−  Check for linear/constant solutions first. They are usually the only ones.
 
−   
−  fff trick is pro
 
−   
−  Pointwise trap?!?
 
−   
−  Be aware of your domain/range. (<math>N, Q, Z, C</math>(Probably not))
 