Trivial Inequality

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 non-negative. 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

Triangle has and . What is the largest area that this triangle can have?
(1992 AIME, Problem 13)
- 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 $9\cdot\frac{1}{2} \cdot \frac{40*41}{9} = 820$ .

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Trivial Inequality

Contents

Applications

Problems

Intermediate

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