Difference between revisions of "Trivial Inequality"

Revision as of 20:44, 22 November 2007

The Trivial Inequality is a simple inequality named because of its sheer simplicity and seeming triviality.

Inequality

The 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.

Proof

We assume the negation of the theorem; that is there is a real $x$ such that $x^2<0$ . Since $x\in \mathbb R$ , and $\{x|x=0\},\{x|x<0\},\{x|x>0\}$ are partitions of $\mathbb R$ , there are three cases for $x$ .

Case 1: $x=0$ : This obviously is a contradiction, as $0^2\not < 0$

Case 2: $x>0$ : Here, we divide by $x$ , which is allowable because we know $x$ is positive: $x<\frac{0}{x}\Rightarrow x<0$ , which results in contradiction.

Case 3: $x<0$ : Since $x<0$ , we can again divide by $x$ and reverse the inequality symbol: $x>\frac{0}{x}\Rightarrow x>0$ , which again is a contradiction.

Thus, the theorem is true by contradiction.

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 5: / Line 5: @@
 ==Proof==
-We assume the negation of the theorem; that is there is a real <math>x</math> such that <math>x^2<0</math>. We now consider three cases:
+We assume the negation of the theorem; that is there is a real <math>x</math> such that <math>x^2<0</math>. Since <math>x\in \mathbb R</math>, and <math>\{x|x=0\},\{x|x<0\},\{x|x>0\}</math> are [[partition]]s of <math>\mathbb R</math>, there are three cases for <math>x</math>.
-'''Case 1; <math>x=0</math>:''' This obviously is a contradiction, as <math>0^2\not < 0</math>
+'''Case 1: <math>x=0</math>:''' This obviously is a contradiction, as <math>0^2\not < 0</math>
-'''Case 2; <math>x>0</math>:''' Here, we divide by <math>x</math>, which is allowable because we know <math>x</math> is positive: <math>x<\frac{0}{x}\Rightarrow x<0</math>, which results in contradiction.
+'''Case 2: <math>x>0</math>:''' Here, we divide by <math>x</math>, which is allowable because we know <math>x</math> is positive: <math>x<\frac{0}{x}\Rightarrow x<0</math>, which results in contradiction.
-'''Case 3; <math>x<0</math>:''' Since <math>x<0</math>, we can again divide by <math>x</math> and reverse the inequality symbol: <math>x>\frac{0}{x}\Rightarrow x>0</math>, which again is a contradiction.
+'''Case 3: <math>x<0</math>:''' Since <math>x<0</math>, we can again divide by <math>x</math> and reverse the inequality symbol: <math>x>\frac{0}{x}\Rightarrow x>0</math>, which again is a contradiction.
-Thus, the theorem is true.
+Thus, the theorem is true by contradiction.
 ==Applications==

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