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Revision as of 21:03, 20 June 2006

The trivial inequality states that ${x^2 \ge 0}$ 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 1992/13) 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 $9\cdot\frac{1}{2} \cdot \frac{40*41}{9} = 820$. $\Box$

Solution credit to: 4everwise


See also

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