Trivial Inequality: Difference between revisions
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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. | 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. | ||
==Applications== | ==Applications== | ||
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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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== Intermediate == | |||
=== AIME 1992, Problem 13 === | |||
''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?'' | ''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?'' | ||
Revision as of 00:49, 18 June 2006
The Inequality
The trivial inequality states that 
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.
Applications
Maximizing and minimizing quadratic functions
After Completing the square, the trivial inequality can be applied to determine the extrema of a quadratic function.
Intermediate
AIME 1992, Problem 13
Triangle 
has 
and 
. What's 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 credit to: 4everwise
