Young's inequality for scalar multiplied with absolute value of function

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Young's inequality states the following:

For scalar time functions $x(t)\in\mathbb{R}$ and $y(t)\in\mathbb{R}$, the following holds:

$$ x\cdot y \leq \frac{1}{2\cdot \epsilon}\cdot x^2 + \frac{\epsilon}{2}\cdot y^2 \ \ \ (1) $$

for any $\epsilon > 0$. Consider now a constant $c>0$ and the absolute value of a function, i.e., $|x(t)|$. Does the inequality also hold ?

$$ c\cdot |x| \leq \frac{1}{2\cdot \epsilon}\cdot c^2 + \frac{\epsilon}{2}\cdot |x|^2 \ \ \ (2)$$

for any $\epsilon > 0$. I am thinking that the constant can be considered as a function in the form:

$$ y(t) = c > 0, \ \forall t \geq 0 $$

and if this is true, then I suppose that Young's inequality $(2)$ also holds.

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For $x,y \in \mathbb{R}$, $|x|^2 = x^2$ and $|y|^2 = y^2$. Therefore $$x\cdot y \leq |x|\cdot |y| \leq \frac{1}{2\cdot \epsilon}\cdot x^2 + \frac{\epsilon}{2}\cdot y^2$$ which means, $$c\cdot |x| \leq |c|\cdot |x| \leq \frac{1}{2\cdot \epsilon}\cdot c^2 + \frac{\epsilon}{2}\cdot x^2$$