Difference between revisions of "Trivial Inequality"
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− | The ''' | + | 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== | ==Proof== | ||
− | We | + | 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== | |
+ | |||
+ | 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> | |
− | |||
− | [[Category: | + | (Solution [[Problems Collection|here]] see problem 13 solution 1) |
− | [[Category: | + | [[Category:Algebra]] |
+ | [[Category:Inequalities]] |
Latest revision as of 20:20, 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
[hide]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)