On a conceptual question on inequalities.

Question

Prologue: $a > b\implies a^2>b^2$ will hold only when $|a| > | b|$.

So I was messing with this inequality for quite some time

$$\sqrt{8+2x-x^2}>6-3x$$

• First things first I found out that the common domain of this inequality is actually $[-2,4]$.

• when $6-3x <0$ or $x>2$ the inequality holds for $(2,\infty)$

When $6-3x \leq0$, the author squared the inequality or in other words " $a^2>b^2$ followed" which in turn implied that

$$\left|\sqrt{ 8+2x-x^2}\right| > |6-3x|$$ holds in order for "$>$" to remain intact.

How do I be sure about this?

Answer 1

$$\sqrt{8+2x-x^2}>6-3x\tag{*}$$

• First things first I found out that the common domain of this inequality is actually $[-2,4]$.

• when $6-3x <0$ or $x>2$ the inequality holds for $(2,\infty)$

Note that the second bullet point is wrong. As you have observed, $$ 8+2x-x^2\geq 0 $$ if and only if $x\in[-2,4]$. This means that the solution set to ($*$) must be a subset of $S:=[-2,4]$ and one should not bother with any point outside this set. Now there are obviously two cases to consider:

• $x\in[-2,4]$ and $6-3x<0$; namely, $x\in(2,4]$.
• $x\in[-2,4]$ and $6-3x\geq 0$; namely, $x\in[-2,2]$.

For the first case, one can see that any $x\in(2,4]$ is a solution to ($*$).

For the second case, note that you are comparing two nonnegative real numbers and for $a,b\geq 0$,

$a>b$ if and only if $a^2>b^2$.

Thus when $x\in[-2,2]$, ($*$) is equivalent to $$ 8+2x-x^2>(6-3x)^2. $$