# Prove that there exist always c s.t. $x(t)(y(t)+1)\leq c x(t)y(t)$ or give a counter example..

I want to know if the following is true and if so how can i prove it: for some positive real valued functions of variable $t$, $x(t)$ and $y(t)$ do i have: there exist always a constant $c$ such that $$x(t)(y(t)+1)\leq c x(t)y(t)$$ or alternatively under which assumption the aformentioned is in fact true?

Since $x(t)$ is positive, the condition is equivalent to $$y(t)+1\le cy(t),$$ hence to $$c\ge 1+\frac 1{y(t)}.$$ We see that such $c$ exists if and only if $\inf y(t)>0$.
\begin{align} & x(t)(y(t)+1)\leq c x(t)y(t) \\ \iff & y(t)+1 \leq c y(t) \tag{$x(t) > 0$} \\ \iff & \frac1c \le 1-\frac{1}{y(t)+1} \\ \iff & y(t) + 1 \ge \frac{c}{c-1} \\ \iff & y(t) \ge \frac{1}{c-1} \end{align}
But $y$ is not necessarily bounded below (say $y(t) = \dfrac{1}{1+t^2}$), so no such $c$ exists.