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Difference between revisions of "Trivial Inequality"

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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.
  
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Another application will be to minimize/maximize [[quadratics]]. For example,
+
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>
 
<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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*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>. [[Inequality_Introductory_Problem_2|Solution]]
 
*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>.  
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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>.  
  
 
===Intermediate===
 
===Intermediate===

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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