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I have

$$ \left| \frac{f(x)}{\sqrt{g(x)}} \right| \leq 1 \quad\quad (1)$$ for $x>0$, and $f(x)$ for some values of $x$ is positive and for some is negative. Now, when I use the identity $\left|y\right|=\sqrt{y^2}$ in $(1)$, I obtain $$ \frac{f(x)^2}{g (x)}\leq 1 \quad\quad (2).$$ But when I square both sides of $(1)$ directly, I obtain $$ \left|\frac{f(x)^2}{g (x)}\right|\leq 1 \quad\quad (3).$$

But the results of $(2)$ and $(3)$ are not the same. I will be grateful if someone explains where I am going wrong?

katy98
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    The results of $(2)$ and $(3)$ are the same. Indeed, $g$ is positive (otherwise its square root has no sense) – TheSilverDoe Aug 27 '20 at 14:54
  • @TheSilverDoe Then, $ \left| f(x)^2 \right| \leq \left| g (x)\right| $ is equivalent with $f(x)^2-g(x)\leq0$? – katy98 Aug 27 '20 at 15:04
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    Here, yes, because $g$ is positive. You always have $|f^2|=f^2$, and because $g$ is positive, you have $|g|=g$. – TheSilverDoe Aug 27 '20 at 15:05
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    @katy98 $|f(x)^2|=f(x)^2$ but in general $|g(x)| \neq g(x)$, unless $g(x)\ge 0$. – user Aug 27 '20 at 15:06

3 Answers3

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As noticed in the comments, note that since $f(x)^2\ge 0$ and since $g(x)>0$ we have that

$$\left|\frac{f(x)^2}{g (x)}\right|=\frac{f(x)^2}{g (x)}\le 1$$

indeed by definition

$$\left| x \right| = \begin{cases} x \hspace{1cm} x \geq 0 \\ -x \hspace{0.7cm} x < 0 \end{cases}$$

user
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  • As Yves Daust mentioned in his answer, am I allowed to square inside the absolute value? I mean does it make sense to write $(1)$ as $(3)$? is it correct to do so? – katy98 Aug 27 '20 at 15:14
  • Yes since$ f(x)=x^2$ is strictly increasing for $x\ge 0$ we have that $A,B\ge0$ we have that $$|A|\le B \iff A\le B \iff A^2\le B^2$$ – user Aug 27 '20 at 15:20
  • $f(x)$ is not $x^2$. It can be negative for some values of $x$ as I wrote in the text. – katy98 Aug 27 '20 at 15:23
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    @katy98 Sorry I used the same symbol but I was referring to a general property. I mean that since the square function $F(x)=x^2$ in increasing for $x\ge 0$, therefore since $|f(x)| \ge 0$ and $g(x)>0$ we have that $$|f(x)|\le g(x) \iff f(x)^2\le g^2(x)$$ – user Aug 27 '20 at 15:26
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In $(2)$, you are not using $|y|=\sqrt{y^2}$ but $|y|^2=y^2$. And in $(3)$, there is no reason to wrap the square inside an absolute value.

In both cases, you have the same problem,

$$\left(\sqrt{g(x)}\right)^2\ne g(x)$$ as functions. They don't have the same domain.

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You have done everything correct. You should know that, Modulus of a variable always returns the positive value of the variable. so, $|x|=x$ so,in the third or first equation you can easily remove the modulus.The values are same with modulus or without modulus.

So now,th three equations are same

user
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Asif Hassan
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