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| − | ==Introduction==
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| − | 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 :)
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| − | ==Inequalities==
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| − | Yay! I love inequalities. Clever algebraic manipulation+thereoms is all you need. It all comes from experience though...
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| − | ==I.Basics==
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| − | AM-GM, Cauchy (Titu's Lemma as well), Muirhead, and Holder's.
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| − | 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.
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| − | Example 1:
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| − | (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>
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| − | Solution:
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| − | 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
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| − | <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>,
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| − | <math>abc \geq 27</math>
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| − | By AM-GM.
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| − | 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
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| − | <math>(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^3=1</math>, we get
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| − | <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>
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| − | Which is true from AM-GM. We shall now introduce Muirhead's...
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| − | Example 2:
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| − | 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>.
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| − | Solution:
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| − | First we homogenize:
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| − | <math>a^2+b^2+c^2 \geq (a+b+c)(abc)^{1/3}</math>
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| − | Which is true because <math>(2,0,0)</math> majorizes <math>(\frac{4}{3}, \frac{1}{3}, \frac{1}{3})</math>
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| − | Cauchy:
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| − | 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
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| − | <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>
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| − | Prove that <math>\text{max}(a_1, a_2, ... ,a_n) \le 4 \text{min}(a_1, a_2, ... , a_n)</math>.
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| − | Try to solve this on your own! Very cute problem.
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| − | Note that you'll probably only ever need Holder's for <math>3</math> variables...
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| − | Example 3 (2004 usamo/5)
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| − | Let <math>a</math>, <math>b</math>, and <math>c</math> be positive real numbers. Prove that
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| − | <math>(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \ge (a+b+c)^3</math>.
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| − | Solution:
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| − | 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.
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| − | 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>.
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| − | 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.
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| − | ==I.More advanced stuff, learn some calculus==
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| − | You will need to know derivatives for this part. It's actually pretty simple.
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| − | Basically, the derivative of <math>x^n</math> is <math>nx^{n-1}</math>
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| − | 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.
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| − | 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.
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| − | 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>,
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| − | <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>
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| − | 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>,
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| − | <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>
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| − | Problem 2: Using Jensen's and Holder's, solve 2001 IMO/2:
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| − | 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>.
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| − | https://artofproblemsolving.com/wiki/index.php/Category:Olympiad_Inequality_Problems
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| − | ==Function Equations==
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