Difference between revisions of "Trivial Inequality"

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==Statement==
 
==Statement==
For all [[real number]]s <math>x</math>, <math>x^2 \ge 0</math>, with equality if and only if <math>x = 0</math>.
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For all [[real number]]s <math>x</math>, <math>x^2 \ge 0</math>, equality holds if and only if <math>x = 0</math>.
  
 
==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 have a contradiction.
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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.
 
Therefore, <math>x^2 \ge 0</math> for all real <math>x</math>, as claimed.
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== Problems ==
 
== Problems ==
 
 
===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>.
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*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]]
 
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*Show that <math>x^2+y^4\geq 2x+4y^2-4</math> for all real <math> x </math>.
  
 
===Intermediate===
 
===Intermediate===

Revision as of 22:10, 29 May 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.

Statement

For all real numbers $x$, $x^2 \ge 0$, equality holds if and only if $x = 0$.

Proof

We can have either $x=0$, $x>0$, or $x<0$. If $x=0$, then $x^2 = 0^2 \ge 0$. If $x>0$, then $x^2 = (x)(x) > 0$ by the closure of the set of positive numbers under multiplication. Finally, if $x<0$, then $x^2 = (-x)(-x) > 0,$ again by the closure of the set of positive numbers under multiplication.

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

Introductory

  • Find all integer solutions $x,y,z$ of the equation $x^2+5y^2+10z^2=4xy+6yz+2z-1$.
  • Show that $\sum_{k=1}^{n}a_k^2 \geq a_1a_2+a_2a_3+\cdots+a_{n-1}a_n+a_na_1$. Solution
  • Show that $x^2+y^4\geq 2x+4y^2-4$ for all real $x$.

Intermediate

  • Triangle $ABC$ has $AB=9$ and $BC: AC=40: 41$. What is the largest area that this triangle can have? (AIME 1992)


Olympiad