# Difference between revisions of "Trivial Inequality"

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== Problems == | == Problems == | ||

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===Introductory=== | ===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>. | *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>\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^2-4</math> for all real <math> x </math>. | ||

===Intermediate=== | ===Intermediate=== |

## Latest revision as of 15:36, 22 April 2021

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 , , equality holds 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 we 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
- Show that for all real .

### 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)