Proving bounded first derivative

Hint:

Your inequality is true for every $y \in \mathbb{R}$ and

$$\min_{0 < z < \infty} \left(\frac{2C}{z} + \frac{Dz}{2}\right)= 2 \sqrt{CD}$$

Here is a solution with Tyler's expansion: $$ \forall x\in\mathbb{R},\forall h>0,\\\ f(x+h)=f(x)+hf'(x)+\frac{h^2}{2}f''(\eta_1)\text{ ,where $\eta_1\in(x,x+h)$}\\ f(x-h)=f(x)-hf'(x)+\frac{h^2}{2}f''(\eta_2)\text{ ,where $\eta_2\in(x-h,x)$}\\ \Rightarrow f(x+h)-f(x-h)=2hf'(x)+\frac{h^2}{2}[f''(\eta_1)+f''(\eta_2)]\\ \text{Hence, }|2hf'(x)|\leq|f(x+h)-f(x-h)|+|\frac{h^2}{2}[f''(\eta_1)+f''(\eta_2)]|\leq\\ 2\sup|f|+h^2\sup |f''| $$ We deduce that: $$2\sup|f'|\leq\frac{2\sup|f|}{h}+h\sup|f''|$$ Since $\sup|f^{(k)}|$ is a constant. We can let $$h=\sqrt{\dfrac{2\sup|f|}{\sup|f''|}}$$ We deduce that $$\boxed{\sup |f'| \le \sqrt{2}\cdot \sqrt{\sup |f| \cdot \sup|f''|}}$$ The first derivative is bounded since the function and its second derivative are bounded.