Difference between revisions of "Trivial Inequality"

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The '''Trivial Inequality''' states that <math>{x^2 \ge 0}</math> for all [[real number]]s <math>x</math>. This is a rather useful [[inequality]] for proving that certain quantities are [[nonnegative]]. The inequality appears to be obvious and unimportant, but it can be a very powerful problem solving technique.
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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.
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==Statement==
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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>.
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==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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Therefore, <math>x^2 \ge 0</math> for all real <math>x</math>, as claimed.
  
 
==Applications==
 
==Applications==
  
The trivial inequality can be used to maximize and minimize quadratic functions.
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The trivial inequality is one of the most commonly used theorems in mathematics. It is very well-known and does not require proof.
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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:
  
After [[completing the square]], the trivial inequality can be applied to determine the extrema of a quadratic function.
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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.
  
 
== Problems ==
 
== Problems ==
=== Intermediate ===
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===Introductory===
#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? <div style='text-align:right;'>([[1992 AIME Problems/Problem 13|1992 AIME, Problem 13]])</div>
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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>.
#*'''Solution:''' First, consider the triangle in a coordinate system with vertices at <math>(0,0)</math>, <math>(9,0)</math>, and <math>(a,b)</math>.<br>Applying the distance formula, we see that <math>\frac{ \sqrt{a^2 + b^2} }{ \sqrt{ (a-9)^2 + b^2 } } = \frac{40}{41}</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]]
#:We want to maximize <math>b</math>, the height, with <math>9</math> being the base. Simplifying gives <math>-a^2 -\frac{3200}{9}a +1600 = b^2</math>. To maximize <math>b</math>, we want to maximize <math>b^2</math>. '''So if we can write: <math>-(a+n)^2+m=b^2</math> then <math>m</math> is the maximum value for <math>b^2</math>.''' This follows directly from the trivial inequality, because if <math>{x^2 \ge 0}</math> then plugging in <math>a+n</math> for <math>x</math> gives us <math>{(a+n)^2 \ge 0}</math>. So we can keep increasing the left hand side of our earlier equation until <math>{(a+n)^2 = 0}</math>. We can factor <math>-a^2 -\frac{3200}{9}a +1600 = b^2</math> into <math>-(a +\frac{1600}{9})^2 +1600+(\frac{3200}{9})^2 = b^2</math>. We find <math>b</math>, and plug into <math>9\cdot\frac{1}{2} \cdot b</math>. Thus, the area is <math>9\cdot\frac{1}{2} \cdot \frac{40*41}{9} = 820</math>.
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*Show that <math>x^2+y^4\geq 2x+4y^2-4</math> for all real <math> x </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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== See also ==
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===Olympiad===
* [[Optimization]]
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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]])
  
 
[[Category:Inequality]]
 
[[Category:Inequality]]
 
[[Category:Theorems]]
 
[[Category:Theorems]]

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.

Statement

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

Proof

We proceed by contradiction. Suppose there exists a real $x$ such that $x^2<0$. We can have either $x=0$, $x>0$, or $x<0$. If $x=0$, then there is a clear contradiction, as $x^2 = 0^2 \not < 0$. If $x>0$, then $x^2 < 0$ gives $x < \frac{0}{x} = 0$ upon division by $x$ (which is positive), so this case also leads to a contradiction. Finally, if $x<0$, then $x^2 < 0$ gives $x > \frac{0}{x} = 0$ upon division by $x$ (which is negative), and yet again we have a contradiction.

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