Problem: Let $f$ be twice differentiable on $\mathbb{R}$ with $\sup|f(x)|=A$ and $\sup|f''(x)|=C$. Prove that $\sup|f'(x)|\leq \sqrt{2AC}$.
Hint: If $f'(x_0)=b>0$, show that $f'(x_0+t)\geq b-C|t|$. Integrate from $x_0-b/C$ to $x_0+b/C$.
Attempt: I can't understand how to work through the hint, so I tried the following. Let t>0. By Taylor's Theorem, $f'(x_0+t)=\frac{1}{h}[f(x_0+t)-f(x_0)]-tf''(c)$ for some $c\in (x_0+t,x_0+2t)$. Thus \begin{align} |f'(x_0+t)|&= \left|\frac{1}{t}[f(x_0+t)-f(x_0)]-\frac{t}{2}f''(c)\right|\\ &\leq \frac{f(x_0+t)}{t}+\frac{|f(x_0)|}{t}+\frac{t}{2}|f''(c)|\\ &\leq \frac{A}{t}+\frac{A}{t}+\frac{Ct}{2}\\ &= \frac{2A}{t}+\frac{Ct}{2} \end{align} I still need the last expression to be less than or equal to $\sqrt{2AC}$. Any help would be appreciated.
Let's use the hint directly: By the MVT have $f'(x_0+t)-f'(x_0)=tf''(\xi)$ with $\xi$ between $x_0$ and $x_0+t$, hence $|f'(x_0+t)-f'(x_0)|\le C|t|$, i.e., $$ f'(x_0)-C|t|\le f'(x_0+t)\le f'(x_0)-C|t|.$$ Thus for $x_0\in \Bbb R$ and $a> 0$, $$\begin{align} f(x_0+a)-f(x_0-a)&=\int_{-a}^af'(x_0+t)\,\mathrm dt\\ &\ge \int_{-a}^a(f'(x_0)-C\left|t\right|)\,\mathrm dt\\ &=\int_{-a}^0(f'(x_0)+Ct)\,\mathrm dt+\int_{0}^a(f'(x_0)-Ct)\,\mathrm dt\\ &=2a f'(x_0)-a^2C \end{align}$$ and $$\begin{align} f(x_0+a)-f(x_0-a)&=\int_{-a}^af'(x_0+t)\,\mathrm dt\\ &\le \int_{-a}^a(f'(x_0)+C\left|t\right|)\,\mathrm dt\\ &=\int_{-a}^0(f'(x_0)-Ct)\,\mathrm dt+\int_{0}^a(f'(x_0)+Ct)\,\mathrm dt\\ &=2a f'(x_0)+a^2C. \end{align}$$ From the bound for $f$ we know $-2A\le f(x_0+a)-f(x_0-a)\le 2A$ so that ultimately $$\frac{-2A-a^2C}{2a}\le f'(x_0)\le\frac{2A+a^2C}{2a}.$$ For the specific choice $a=\sqrt{\frac{2A}{C}}$ this leads to $$ -\sqrt{ 2AC}\le f'(x_0)\le\sqrt{2AC}$$ as desired.