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Difference between revisions of "Trivial Inequality"

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==Statement==
 
==Statement==
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For all [[real number]]s <math>x</math>, <math>{x^2 \ge 0}</math>.
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For all [[real number]]s <math>x</math>, <math>x^2 \ge 0</math>.
  
 
==Proof==
 
==Proof==
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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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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.
 
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>
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 +
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 ==
 
== Problems ==
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===Introductory===
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*Find all integer solutions <math>x,y,z</math> of the equation <math>x^2+5y^2+10z^2=4xy+6yz+2z-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^2-5</math> for all real <math>x</math> and <math>y</math>.
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===Intermediate===
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*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]])
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# Find all integer solutions <math>x,y,z</math> of the equation <math>x^2+5y^2+10z^2=2z+6yz+4xy-1</math>.
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===Olympiad===
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# ([[1992 AIME Problems/Problem 13|AIME 1992]]) Triangle <math>ABC</math> has <math>AB</math><math>=9</math> and <math>BC: AC=40: 41</math>. What is the largest area that this triangle can have?
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*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]])
  
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[[Category:Inequality]]
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[[Category:Algebra]]
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[[Category:Theorems]]
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[[Category:Inequalities]]

Revision as of 16:47, 27 November 2022

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 $x$, $x^2 \ge 0$.

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.

Another application will be to minimize/maximize quadratics. For example,

\[ax^2+bx+c = a(x^2+\frac{b}{a}x+\frac{b^2}{4a^2})+c-\frac{b^2}{4a} = a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}.\]

Then, we use trivial inequality to get $ax^2+bx+c\ge c-\frac{b^2}{4a}$ if $a$ is positive and $ax^2+bx+c\le c-\frac{b^2}{4a}$ if $a$ is negative.

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 $x^2+y^4\geq 2x+4y^2-5$ for all real $x$ and $y$.

Intermediate

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


Olympiad

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