Unconditional Inequality/Examples/Arbitrary Example 1
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Example of Unconditional Inequality
The inequality:
- $2 x^2 + 1 > x - 1$
is an unconditional inequality.
Proof
\(\ds 2 x^2 + 1\) | \(>\) | \(\ds x - 1\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds 2 x^2 + 1 - x + 1\) | \(>\) | \(\ds 0\) | |||||||||||
\(\text {(1)}: \quad\) | \(\ds \leadstoandfrom \ \ \) | \(\ds 2 x^2 - x + 2\) | \(>\) | \(\ds 0\) |
It remains to be shown that $(1)$ is true for all $x \in \R$.
Differentiating $y = 2 x^2 - x + 2$ with respect to $x$ gives us:
- $\dfrac {\d y} {\d x} = 4 x - 1$
which is increasing throughout and is zero at $x = \dfrac 1 4$.
At $x = \dfrac 1 4$ we have that:
- $y = 2 \times \paren {\dfrac 1 4}^2 - \dfrac 1 4 + 2 = 1 \frac 7 8 > 0$
As the coefficient of $x$ is positive, this means $y$ has a minimum at $x = \dfrac 1 4$.
As that is positive, that means $2 x^2 - x + 2$ never goes negative.
Hence we have that:
- $\forall x \in \R: 2 x^2 + 1 > x - 1$
as we were required to show.
$\blacksquare$
Sources
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): inequality
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): inequality: 2.