Unconditional Inequality/Examples/Arbitrary Example 1

Example of Unconditional Inequality

$2 x^2 + 1 > x - 1$

$\ds 2 x^2 + 1$

$>$

$\ds x - 1$

$\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.

$\blacksquare$