Difference between revisions of "Trivial Inequality"

Latest revision as of 20:20, 2 August 2024

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)

The fraction,

$\frac{ab+bc+ac}{(a+b+c)^2}$

where $a,b$ and $c$ are side lengths of a triangle, lies in the interval $(p,q]$ , where $p$ and $q$ are rational numbers. Then, $p+q$ can be expressed as $\frac{r}{s}$ , where $r$ and $s$ are relatively prime positive integers. Find $r+s$ . (Solution here see problem 3 solution 1)

Olympiad

Let $c$ be the length of the hypotenuse of a right triangle whose two other sides have lengths $a$ and $b$ . Prove that $a+b\le c\sqrt{2}$ . When does the equality hold? (1969 Canadian MO)

Let $x,y$ and $z$ be real numbers. Show that

$(x^2+z^2)^2+y^4 \ge 4xzy^2$

(Solution here see problem 13 solution 1)

@@ Line 1: / Line 1: @@
-==The Inequality==
+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.
-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.
+==Statement==
+For all [[real number]]s <math>x</math>, <math>x^2 \ge 0</math>.
+==Proof==
+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.
 ==Applications==
-'''Maximizing and minimizing quadratic functions'''
+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 [[AMGM | Arithmetic Mean-Geometric Mean]] inequality:
+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>
+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 ==
+===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>.
+*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^2-5</math> for all real <math>x</math> and <math>y</math>.
+===Intermediate===
+*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]])
+*The fraction,
-After [[Completing the square]], the trivial inequality can be applied to determine the extrema of a quadratic function.
+<cmath>\frac{ab+bc+ac}{(a+b+c)^2}</cmath>
-==USA AIME 1992, Problem 13==
+where <math>a,b</math> and <math>c</math> are side lengths of a triangle, lies in the interval <math>(p,q]</math>, where <math>p</math> and <math>q</math> are rational numbers. Then, <math>p+q</math> can be expressed as <math>\frac{r}{s}</math>, where <math>r</math> and <math>s</math> are relatively prime positive integers. Find <math>r+s</math>. (Solution [[Problems Collection|here]] see problem 3 solution 1)
-''Triangle <math>ABC</math> has <math>AB</math><math>=9</math> and <math>BC: AC=40: 41</math>. What's the largest area that this triangle can have?''
+===Olympiad===
+*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]])
-Solution:
+*Let <math>x,y</math> and <math>z</math> be real numbers. Show that
-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>.
-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>.
+<cmath>(x^2+z^2)^2+y^4 \ge 4xzy^2</cmath>
-''Solution credit to: 4everwise''
+(Solution [[Problems Collection|here]] see problem 13 solution 1)
+[[Category:Algebra]]
+[[Category:Inequalities]]

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