Difference between revisions of "Trivial Inequality"

Revision as of 02:45, 15 March 2008

The trivial inequality is an inequality that states that the square of any real number is nonnegative. Its name comes from its simplicity and straightforwardness.

Statement

For all real numbers $x$ , ${x^2 \ge 0}$ .

Proof

We proceed by contradiction. Suppose there exists a real $x$ such that $x^2<0$ . We can have either $x=0$ , $x>0$ , or $x<0$ . If $x=0$ , then there is a clear contradiction, as $x^2 = 0^2 \not < 0$ . If $x>0$ , then $x^2 < 0$ gives $x < \frac{0}{x} = 0$ upon division by $x$ (which is positive), so this case also leads to a contradiction. Finally, if $x<0$ , then $x^2 < 0$ gives $x > \frac{0}{x} = 0$ upon division by $x$ (which is negative), and yet again we have a contradiction.

Therefore, $x^2 \ge 0$ for all real $x$ , as claimed.

Applications

The trivial inequality is one of the most commonly used theorems in mathematics. It is very well-known and does not require proof.

One application is maximizing and minimizing quadratic functions. It gives an easy proof of the two-variable case of the Arithmetic Mean-Geometric Mean inequality:

Suppose that $x$ and $y$ are nonnegative reals. By the trivial inequality, we have $(x-y)^2 \geq 0$ , or $x^2-2xy+y^2 \geq 0$ . Adding $4xy$ to both sides, we get $x^2+2xy+y^2 = (x+y)^2 \geq 4xy$ . Since both sides of the inequality are nonnegative, it is equivalent to $x+y \ge 2\sqrt{xy}$ , and thus we have $\frac{x+y}{2} \geq \sqrt{xy},$ as desired.

Problems

Find all integer solutions $x,y,z$ of the equation $x^2+5y^2+10z^2=2z+6yz+4xy-1$ .
(AIME 1992) Triangle $ABC$ has $AB$ $=9$ and $BC: AC=40: 41$ . What is the largest area that this triangle can have?

@@ Line 1: / Line 1: @@
-The '''trivial inequality''' is a simple [[inequality]] named because of its sheer simplicity and seeming triviality.
+The '''trivial inequality''' is an [[inequality]] that states that the square of any real number is nonnegative. Its name comes from its simplicity and straightforwardness.
-==Inequality==
+==Statement==
-The inequality states that <math>{x^2 \ge 0}</math> for all [[real number]]s <math>x</math>. 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.
+For all [[real number]]s <math>x</math>, <math>{x^2 \ge 0}</math>.
 ==Proof==
-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>.
+We proceed by contradiction. Suppose there exists a real <math>x</math> such that <math>x^2<0</math>. We can have either <math>x=0</math>, <math>x>0</math>, or <math>x<0</math>. If <math>x=0</math>, then there is a clear contradiction, as <math>x^2 = 0^2 \not < 0</math>. If <math>x>0</math>, then <math>x^2 < 0</math> gives <math>x < \frac{0}{x} = 0</math> upon division by <math>x</math> (which is positive), so this case also leads to a contradiction. Finally, if <math>x<0</math>, then <math>x^2 < 0</math> gives <math>x > \frac{0}{x} = 0</math> upon division by <math>x</math> (which is negative), and yet again we have a contradiction.
-'''Case 1: <math>x=0</math>:''' This obviously is a contradiction, as <math>0^2\not < 0</math>
+Therefore, <math>x^2 \ge 0</math> for all real <math>x</math>, as claimed.
-'''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.
-Thus, the theorem is true by contradiction.
 ==Applications==
-The trivial inequality can be used to maximize and minimize [[quadratic]] functions.
+The trivial inequality is one of the most commonly used theorems in mathematics. It is very well-known and does not require proof.
-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:
+One application is maximizing and minimizing [[quadratic]] functions. It gives an easy proof of the two-variable case of the [[AMGM | Arithmetic Mean-Geometric Mean]] inequality:
-Suppose that <math>a,b</math> are nonnegative [[real number]]s. Starting with <math>(a-b)^2\geq0</math>, after squaring we have <math>a^2-2ab+b^2\geq0</math>. Now add <math>4ab</math> to both sides of the inequality to get <math>a^2+2ab+b^2=(a+b)^2\geq4ab</math>. If we take the square root of both sides (since both sides are nonnegative) and divide by 2, we have the well-known [[AMGM | Arithmetic Mean-Geometric Mean]] Inequality for 2 variables: <math>\frac{a+b}2\geq\sqrt{ab}</math>
+Suppose that <math>x</math> and <math>y</math> are nonnegative reals. By the trivial inequality, we have <math>(x-y)^2 \geq 0</math>, or <math>x^2-2xy+y^2 \geq 0</math>. Adding <math>4xy</math> to both sides, we get <math>x^2+2xy+y^2 = (x+y)^2 \geq 4xy</math>. Since both sides of the inequality are nonnegative, it is equivalent to <math>x+y \ge 2\sqrt{xy}</math>, and thus we have <cmath> \frac{x+y}{2} \geq \sqrt{xy}, </cmath> as desired.
 == Problems ==
-=== 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>. (Hint: rewrite as <math>(x-2y)^2+(y-3z)^2+(z-1)^2=0</math>)
-=== Intermediate ===
+# Find all integer solutions <math>x,y,z</math> of the equation <math>x^2+5y^2+10z^2=2z+6yz+4xy-1</math>.
-*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]])
+# ([[1992 AIME Problems/Problem 13|AIME 1992]]) 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?
 == See also ==

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