The question comes from reading the question and the first answer here.
Let us consider a specific example, \begin{equation} y'(x)\le \frac{1}{2}y^2(x), \quad 1>x\ge 0 \end{equation} which satisfies the conditions of Bihara's theorem (differential form). Therefore, one can obtain \begin{equation} y\le -\frac{2}{x}, \end{equation} i.e., by substituting $y(0)=-\infty$ and $C=1/2$ in @Lutz Lehmann's answer.
In contrast, let us check the result $y\le -\frac{2}{x}$. we know that \begin{equation} -\frac{2}{x^2}\le -\frac{2}{x}, \quad \text{when}\quad 1>x\ge 0, \end{equation} thus I expect that $y=-\frac{2}{x^2}$ must satisfy the initial inequality. However, if we substitute it into $y'(x)\le \frac{1}{2}y^2(x)$, we get \begin{equation} x\le 1/2, \end{equation} which of course contradicts to $y'(x)\le \frac{1}{2}y^2(x)$ when $x\in (1/2,1)$.
So what's the problem here?