Difference between revisions of "Trivial Inequality"
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After [[completing the square]], the trivial inequality can be applied to determine the extrema of a quadratic function. | After [[completing the square]], the trivial inequality can be applied to determine the extrema of a quadratic function. | ||
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+ | Here is an example of the important use of this inequality: | ||
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+ | Suppose that <math>a,b</math> are nonnegative [[real number]]s. Starting with <math>(a-b)^2\geq0</math>, after squaring we have <math>a^2-2ab+b^2\geq0</math>. Now add <math>4ab</math> to both sides of the inequality to get <math>a^2+2ab+b^2=(a+b)^2\geq4ab</math>. If we take the square root of both sides (since both sides are nonnegative) and divide by 2, we have the well-known [[AMGM | Arithmetic Mean-Geometric Mean]] Inequality for 2 variables: <math>\frac{a+b}2\geq\sqrt{ab}</math> | ||
== Problems == | == Problems == |
Revision as of 12:26, 15 November 2007
The Trivial Inequality states that for all real numbers . This is a rather useful inequality for proving that certain quantities are nonnegative. The inequality appears to be obvious and unimportant, but it can be a very powerful problem solving technique.
Contents
[hide]Applications
The trivial inequality can be used to maximize and minimize quadratic functions.
After completing the square, the trivial inequality can be applied to determine the extrema of a quadratic function.
Here is an example of the important use of this inequality:
Suppose that are nonnegative real numbers. Starting with , after squaring we have . Now add to both sides of the inequality to get . If we take the square root of both sides (since both sides are nonnegative) and divide by 2, we have the well-known Arithmetic Mean-Geometric Mean Inequality for 2 variables:
Problems
Intermediate
- Triangle has and . What is the largest area that this triangle can have?
- Solution: First, consider the triangle in a coordinate system with vertices at , , and .
Applying the distance formula, we see that .
- We want to maximize , the height, with being the base. Simplifying gives . To maximize , we want to maximize . So if we can write: then is the maximum value for . This follows directly from the trivial inequality, because if then plugging in for gives us . So we can keep increasing the left hand side of our earlier equation until . We can factor into . We find , and plug into . Thus, the area is .
- Solution: First, consider the triangle in a coordinate system with vertices at , , and .