Difference between revisions of "Trivial Inequality"

Revision as of 18:14, 15 November 2007

The Trivial Inequality states that ${x^2 \ge 0}$ for all real numbers $x$ . This is a rather useful inequality for proving that certain quantities are nonnegative. The inequality appears to be obvious and unimportant, but it can be a very powerful problem solving technique.

Applications

The trivial inequality can be used to maximize and minimize quadratic functions.

After completing the square, the trivial inequality can be applied to determine the extrema of a quadratic function.

Here is an example of the important use of this inequality:

Suppose that $a,b$ are nonnegative real numbers. Starting with $(a-b)^2\geq0$ , after squaring we have $a^2-2ab+b^2\geq0$ . Now add $4ab$ to both sides of the inequality to get $a^2+2ab+b^2=(a+b)^2\geq4ab$ . If we take the square root of both sides (since both sides are nonnegative) and divide by 2, we have the well-known Arithmetic Mean-Geometric Mean Inequality for 2 variables: $\frac{a+b}2\geq\sqrt{ab}$

Problems

Introductory

Find all integer solutions $x,y,z$ of the equation $x^2+5y^2+10z^2=2z+6yz+4xy-1$ . (No source nor solution)

Intermediate

Triangle $ABC$ has $AB$ $=9$ and $BC: AC=40: 41$ . What is the largest area that this triangle can have? (Source)

@@ Line 12: / Line 12: @@
 == Problems ==
-=== Warm-up ===
+=== Introductory  ===
-#Find all integer solutions <math>x,y,z</math> of the equation <math>x^2+5y^2+10z^2=2z+6yz+4xy-1</math>.
+*Find all integer solutions <math>x,y,z</math> of the equation <math>x^2+5y^2+10z^2=2z+6yz+4xy-1</math>. (No source nor solution)
-#*'''Solution:''' ''Challenge to the reader --[[User:10000th User|10000th User]] 12:58, 15 November 2007 (EST)''
 === Intermediate ===
-#Triangle <math>ABC</math> has <math>AB</math><math>=9</math> and <math>BC: AC=40: 41</math>. What is the largest area that this triangle can have? <div style='text-align:right;'>([[1992 AIME Problems/Problem 13|1992 AIME, Problem 13]])</div>
+*Triangle <math>ABC</math> has <math>AB</math><math>=9</math> and <math>BC: AC=40: 41</math>. What is the largest area that this triangle can have? ([[1992 AIME Problems/Problem 13|Source]])
-#*'''Solution:''' First, consider the triangle in a coordinate system with vertices at <math>(0,0)</math>, <math>(9,0)</math>, and <math>(a,b)</math>.<br>Applying the distance formula, we see that <math>\frac{ \sqrt{a^2 + b^2} }{ \sqrt{ (a-9)^2 + b^2 } } = \frac{40}{41}</math>.
-#:We want to maximize <math>b</math>, the height, with <math>9</math> being the base. Simplifying gives <math>-a^2 -\frac{3200}{9}a +1600 = b^2</math>. To maximize <math>b</math>, we want to maximize <math>b^2</math>. '''So if we can write: <math>-(a+n)^2+m=b^2</math> then <math>m</math> is the maximum value for <math>b^2</math>.''' This follows directly from the trivial inequality, because if <math>{x^2 \ge 0}</math> then plugging in <math>a+n</math> for <math>x</math> gives us <math>{(a+n)^2 \ge 0}</math>. So we can keep increasing the left hand side of our earlier equation until <math>{(a+n)^2 = 0}</math>. We can factor <math>-a^2 -\frac{3200}{9}a +1600 = b^2</math> into <math>-(a +\frac{1600}{9})^2 +1600+(\frac{3200}{9})^2 = b^2</math>. We find <math>b</math>, and plug into <math>9\cdot\frac{1}{2} \cdot b</math>. Thus, the area is <math>9\cdot\frac{1}{2} \cdot \frac{40*41}{9} = 820</math>.
 == See also ==

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