Difference between revisions of "Trivial Inequality"

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The '''trivial inequality''' states that <math>{x^2 \ge 0}</math> for all x. This is a rather useful inequality for proving that certain quantities are non-negative. 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>.
  
==Applications==
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==Proof==
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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.
  
'''Maximizing and minimizing quadratic functions'''
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Therefore, <math>x^2 \ge 0</math> for all real <math>x</math>, as claimed.
  
After [[Completing the square]], the trivial inequality can be applied to determine the extrema of a quadratic function.
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==Applications==
  
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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.
  
== Intermediate ==
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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:
  
([[AIME]] 1992/13) 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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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.
  
'''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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== 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^2-4</math> for all real <math> x </math>.  
  
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>. <math>\Box</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]])
  
''Solution credit to: 4everwise''
 
  
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===Olympiad===
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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]])
  
== See also ==
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[[Category:Inequality]]
* [[inequalities]]
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[[Category:Theorems]]
* [[optimization]]
 

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