Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Trivial Inequality"

(fmt)
m
Line 1: Line 1:
−
The '''trivial inequality''' states that <math>{x^2 \ge 0}</math> for all real numbers <math>x</math>. 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 [[real number]]s <math>x</math>. 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.
  
 
==Applications==
 
==Applications==
Line 14: Line 14:
  
 
== See also ==
 
== See also ==
−
* [[Inequalities]]
 
 
* [[Optimization]]
 
* [[Optimization]]

Revision as of 15:45, 18 March 2007

The Trivial Inequality states that ${x^2 \ge 0}$ for all real numbers $x$. 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.

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.

Problems

Intermediate

  1. Triangle $ABC$ has $AB$$=9$ and $BC: AC=40: 41$. What is the largest area that this triangle can have?
    • Solution: First, consider the triangle in a coordinate system with vertices at $(0,0)$, $(9,0)$, and $(a,b)$.
      Applying the distance formula, we see that $\frac{ \sqrt{a^2 + b^2} }{ \sqrt{ (a-9)^2 + b^2 } } = \frac{40}{41}$.
    We want to maximize $b$, the height, with $9$ being the base. Simplifying gives $-a^2 -\frac{3200}{9}a +1600 = b^2$. To maximize $b$, we want to maximize $b^2$. So if we can write: $-(a+n)^2+m=b^2$ then $m$ is the maximum value for $b^2$. This follows directly from the trivial inequality, because if ${x^2 \ge 0}$ then plugging in $a+n$ for $x$ gives us ${(a+n)^2 \ge 0}$. So we can keep increasing the left hand side of our earlier equation until ${(a+n)^2 = 0}$. We can factor $-a^2 -\frac{3200}{9}a +1600 = b^2$ into $-(a +\frac{1600}{9})^2 +1600+(\frac{3200}{9})^2 = b^2$. We find $b$, and plug into $9\cdot\frac{1}{2} \cdot b$. Thus, the area is $9\cdot\frac{1}{2} \cdot \frac{40*41}{9} = 820$.

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Trivial_Inequality&oldid=14106"