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, | ||
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+ | <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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+ | 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 == |
Revision as of 23:20, 6 August 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.
Contents
[hide]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 .
Intermediate
- Triangle has and . What is the largest area that this triangle can have? (AIME 1992)
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)