Difference between revisions of "Trivial Inequality"
(→Proof) |
m (→Intermediate) |
||
(6 intermediate revisions by 4 users not shown) | |||
Line 2: | Line 2: | ||
==Statement== | ==Statement== | ||
− | For all [[real number]]s <math>x</math>, <math>x^2 \ge | + | For all [[real number]]s <math>x</math>, <math>x^2 \ge 0</math>. |
==Proof== | ==Proof== | ||
Line 16: | Line 16: | ||
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> | ||
+ | |||
+ | 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 == | ||
Line 21: | Line 27: | ||
*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]] | ||
− | *Show that <math>x^2+y^4\geq 2x+4y^2- | + | *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=== | ||
− | *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]]) | + | *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, | ||
+ | |||
+ | <math>\frac{ab+bc+ac}{(a+b+c)^2}</math> | ||
+ | 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 [[User:Ddk001#Solution_1.28Probably_official_MAA.2C_lots_of_proofs.29|here]]) | ||
===Olympiad=== | ===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]]) | *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: | + | [[Category:Algebra]] |
− | [[Category: | + | [[Category:Inequalities]] |
Latest revision as of 22:51, 13 January 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.
Contents
Statement
For all real numbers ,
.
Proof
We can have either ,
, or
. If
, then
. If
, then
by the closure of the set of positive numbers under multiplication. Finally, if
, then
again by the closure of the set of positive numbers under multiplication.
Therefore, for all real
, 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 and
are nonnegative reals. By the trivial inequality, we have
, or
. Adding
to both sides, we get
. Since both sides of the inequality are nonnegative, it is equivalent to
, and thus we have
as desired.
Another application will be to minimize/maximize quadratics. For example,
Then, we use trivial inequality to get if
is positive and
if
is negative.
Problems
Introductory
- Find all integer solutions
of the equation
.
- Show that
. Solution
- Show that
for all real
and
.
Intermediate
- Triangle
has
and
. What is the largest area that this triangle can have? (AIME 1992)
- The fraction,
where and
are side lengths of a triangle, lies in the interval
, where
and
are rational numbers. Then,
can be expressed as
, where
and
are relatively prime positive integers. Find
. (Solution here)
Olympiad
- Let
be the length of the hypotenuse of a right triangle whose two other sides have lengths
and
. Prove that
. When does the equality hold? (1969 Canadian MO)