Difference between revisions of "Trivial Inequality"
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− | + | 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 number]]s <math>x</math>, <math>x^2 \ge 0</math>. | ||
+ | |||
+ | ==Proof== | ||
+ | We can have either <math>x=0</math>, <math>x>0</math>, or <math>x<0</math>. If <math>x=0</math>, then <math>x^2 = 0^2 \ge 0</math>. If <math>x>0</math>, then <math>x^2 = (x)(x) > 0</math> by the closure of the set of positive numbers under multiplication. Finally, if <math>x<0</math>, then <math>x^2 = (-x)(-x) > 0,</math> again by the closure of the set of positive numbers under multiplication. | ||
+ | |||
+ | Therefore, <math>x^2 \ge 0</math> for all real <math>x</math>, as claimed. | ||
==Applications== | ==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 [[AMGM | Arithmetic Mean-Geometric Mean]] inequality: | ||
+ | |||
+ | 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. | ||
+ | |||
+ | Another application will be to minimize/maximize quadratics. For example, | ||
+ | |||
+ | <cmath>ax^2+bx+c = a(x^2+\frac{b}{a}x+\frac{b^2}{4a^2})+c-\frac{b^2}{4a} = a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}.</cmath> | ||
+ | |||
+ | Then, we use trivial inequality to get <math>ax^2+bx+c\ge c-\frac{b^2}{4a}</math> if <math>a</math> is positive and <math>ax^2+bx+c\le c-\frac{b^2}{4a}</math> if <math>a</math> is negative. | ||
+ | |||
+ | == Problems == | ||
+ | ===Introductory=== | ||
+ | *Find all integer solutions <math>x,y,z</math> of the equation <math>x^2+5y^2+10z^2=4xy+6yz+2z-1</math>. | ||
+ | *Show that <math>\sum_{k=1}^{n}a_k^2 \geq a_1a_2+a_2a_3+\cdots+a_{n-1}a_n+a_na_1</math>. [[Inequality_Introductory_Problem_2|Solution]] | ||
+ | *Show that <math>x^2+y^4\geq 2x+4y^2-5</math> for all real <math>x</math> and <math>y</math>. | ||
+ | |||
+ | ===Intermediate=== | ||
+ | *Triangle <math>ABC</math> has <math>AB=9</math> and <math>BC: AC=40: 41</math>. What is the largest area that this triangle can have? ([[1992 AIME Problems/Problem 13|AIME 1992]]) | ||
+ | |||
+ | *The fraction, | ||
− | + | <cmath>\frac{ab+bc+ac}{(a+b+c)^2}</cmath> | |
− | + | where <math>a,b</math> and <math>c</math> are side lengths of a triangle, lies in the interval <math>(p,q]</math>, where <math>p</math> and <math>q</math> are rational numbers. Then, <math>p+q</math> can be expressed as <math>\frac{r}{s}</math>, where <math>r</math> and <math>s</math> are relatively prime positive integers. Find <math>r+s</math>. (Solution [[Problems Collection|here]] see problem 3 solution 1) | |
− | + | ===Olympiad=== | |
+ | *Let <math>c</math> be the length of the [[hypotenuse]] of a [[right triangle]] whose two other sides have lengths <math>a</math> and <math>b</math>. Prove that <math>a+b\le c\sqrt{2}</math>. When does the equality hold? ([[1969 Canadian MO Problems/Problem 3|1969 Canadian MO]]) | ||
− | + | *Let <math>x,y</math> and <math>z</math> be real numbers. Show that | |
− | |||
− | + | <cmath>(x^2+z^2)^2+y^4 \ge 4xzy^2</cmath> | |
− | + | (Solution [[Problems Collection|here]] see problem 13 solution 1) | |
+ | [[Category:Algebra]] | ||
+ | [[Category:Inequalities]] |
Latest revision as of 21:20, 2 August 2024
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 , .
Proof
We can have either , , or . If , then . If , then by the closure of the set of positive numbers under multiplication. Finally, if , then again by the closure of the set of positive numbers under multiplication.
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.
Another application will be to minimize/maximize quadratics. For example,
Then, we use trivial inequality to get if is positive and if is negative.
Problems
Introductory
- Find all integer solutions of the equation .
- Show that . Solution
- Show that for all real and .
Intermediate
- Triangle has and . What is the largest area that this triangle can have? (AIME 1992)
- The fraction,
where and are side lengths of a triangle, lies in the interval , where and are rational numbers. Then, can be expressed as , where and are relatively prime positive integers. Find . (Solution here see problem 3 solution 1)
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)
- Let and be real numbers. Show that
(Solution here see problem 13 solution 1)