Differentiation doesn't work here

Assume that $x=I+f$ where $I$ is integral part and $f$ is fractional part.

Putting this in the given inequality it becomes

$I^2-I(I+f)+4 \leq 0 \\ If \geq 4$

For minimum $x$ we have $If=4 \\ f=\dfrac{4}{I}$ .

The smallest $I$ for which $f$ will be a fraction is $I=5$

So, $I=5$ and $f=\dfrac{4}{5}$

Then $x=5+ \dfrac{4}{5}=5.8$

Using the fractional part $\{x\} := x - \lfloor x \rfloor$ , we can rewrite the inequality as $4 \leq \{x\}\cdot\lfloor x \rfloor$

Since $\{x\} \geq 0$ , this inequality implies $\lfloor{x}\rfloor > 0$ and $\{x\} > 0$ . Also, since $\lfloor{x}\rfloor > 0$ and $\{x\} < 1$ , we have $4 \leq \{x\}\cdot\lfloor x \rfloor < \lfloor{x}\rfloor,$ so $\lfloor{x}\rfloor \geq 5$ , and therefore we may consider $\lfloor{x}\rfloor = 5$ . Then $4 \leq \{x\} \cdot 5 \,\,\,\,\, \Rightarrow \,\,\,\,\, \{x\} \geq \frac{4}{5}$ Therefore, the smallest $x$ will have $\{x\} = \frac{4}{5}$ , and $x = \lfloor{x}\rfloor + \{x\} \geq 5 + \frac{4}{5} = 5.8$

By direct evaluation, we can confirm that $x=5.8$ satisfies the inequality, so we have confirmed it's the smallest real number satisfying the inequality.