A Normalization Trick for Inequalities

If you’ve ever taught (or learned) inequalities, you know the usual toolbox: sign charts, tables of intervals, discriminants, and “test points”. 

In my draft on the Unified Substitution Method for Integration (USM), I ended up using a very systematic normalization of quadratics and radicals: you shift and scale so that everything happens in the nice, symmetric interval $[-1,1]$ or outside it.

That same normalization idea leads to a neat side trick for solving inequalities. This post presents that trick, shows it on two main examples, and compares it to the traditional methods.

We’ll use three inequalities:

1. A rational inequality:

$$\frac{x-2}{x+3} \ge 0$$

2. A quadratic that changes sign:

$$x^2 - 4x + 3 \le 0$$

3. A quadratic that is always positive:

$$x^2 - 2x + 3 \ge 0$$

This is basically the same normalization that USM uses for integrals, but repurposed for inequalities.

The core idea: normalize to $[-1,1]$

The general pattern is:

• Rewrite the expression so that it involves

$$\frac{(x+b)-a}{(x+b)+a}

\quad\text{or}\quad

(x+b)^2 \pm a^2$$

with $a>0$.

• Notice that

$$(x+b)-a = a\Big(\frac{x+b}{a}-1\Big),\quad

(x+b)+a = a\Big(\frac{x+b}{a}+1\Big),$$

so the sign is controlled by how $\dfrac{x+b}{a}$ compares to $-1$ and $1$.

• Reduce the problem to very simple linear inequalities in $\dfrac{x+b}{a}$.

Example 1 – Rational inequality

$$\frac{x-2}{x+3} \ge 0$$

Traditional method (very briefly)

Zeros and pole:

$x-2=0 \Rightarrow x=2$,
$x+3=0 \Rightarrow x=-3$.

A sign chart on the intervals $(-\infty,-3)$, $(-3,2)$, $(2,\infty)$ shows the fraction is $\ge 0$ on

$$(-\infty,-3)\cup[2,\infty).$$

Normalization trick

Write

$$\frac{x-2}{x+3} = \frac{x+\color{red}{b-a}}{x+\color{red}{b+a}}$$

and solve this little system:

$$\left. b-a=-2\atop a+b=3\right\}$$

This gives:

$$a = \frac{5}{2},\quad b = \frac{1}{2}.$$

Now plug $a=\dfrac52$, $b=\dfrac12$:

• $\dfrac{x+b}{a} \ge 1 \Rightarrow x \ge 2$,
• $\dfrac{x+b}{a} \le -1 \Rightarrow x \le -3$, but $x=-3$ is excluded.

So again:

$$(-\infty,-3)\cup[2,\infty).$$

No sign chart; just a tiny $2\times 2$ system and linear inequalities.

Example 2 – Quadratic inequality

$$x^2 - 11x + 13 \ge 0$$

Traditional method (very briefly)

Compute the discriminant:

$$\Delta = 11^2 - 4\cdot 1 \cdot 13 = 121 - 52 = 69 > 0.$$

Roots:

$$x = \frac{11 \pm \sqrt{69}}{2}.$$

Since the parabola opens upwards, the quadratic is $\ge 0$ outside the interval between the roots:

$$x \in \left(-\infty,\frac{11-\sqrt{69}}{2}\right] \cup \left[\frac{11+\sqrt{69}}{2},\infty\right).$$

Normalization trick (completing the square)

Complete the square:

$$x^2 - 11x + 13 = \left(x - \frac{11}{2}\right)^2 - \frac{69}{4}.$$

This fits

$$(x+\color{red}{b})^2 - \color{red}{a^2}$$

with

$$b = -\frac{11}{2}, \quad a = \frac{\sqrt{69}}{2}.$$

Again, we have:

$$\frac{x+b}{a} \le -1 \quad\text{or}\quad \frac{x+b}{a} \ge 1.$$

Substituting $b=-\dfrac{11}{2}$, $a=\dfrac{\sqrt{69}}{2}$:

$$x \le \frac{11-\sqrt{69}}{2}\quad\text{or}\quad x \ge \frac{11+\sqrt{69}}{2}.$$

So again,

$$x \in \left(-\infty,\frac{11-\sqrt{69}}{2}\right] \cup \left[\frac{11+\sqrt{69}}{2},\infty\right).$$

To be continued...