Difference between revisions of "Trivial Inequality"
(→Introductory) |
Justin2000 (talk | contribs) |
||
Line 5: | Line 5: | ||
==Proof== | ==Proof== | ||
− | 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 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 Sex have a contradiction. |
Therefore, <math>x^2 \ge 0</math> for all real <math>x</math>, as claimed. | Therefore, <math>x^2 \ge 0</math> for all real <math>x</math>, as claimed. |
Revision as of 19:31, 17 June 2014
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.
Contents
Statement
For all real numbers , , with equality if and only if .
Proof
We proceed by contradiction. Suppose there exists a real such that . We can have either , , or . If , then there is a clear contradiction, as . If , then gives upon division by (which is positive), so this case also leads to a contradiction. Finally, if , then gives upon division by (which is negative), and yet again Sex have a contradiction.
Therefore, for all real , 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 and are nonnegative reals. By the trivial inequality, we have , or . Adding to both sides, we get . Since both sides of the inequality are nonnegative, it is equivalent to , and thus we have as desired.
Problems
Introductory
- Find all integer solutions of the equation .
- Show that . Solution
Intermediate
- Triangle has and . What is the largest area that this triangle can have? (AIME 1992)
Olympiad
- Let be the length of the hypotenuse of a right triangle whose two other sides have lengths and . Prove that . When does the equality hold? (1969 Canadian MO)