Let $f:\mathbb{R}\to\mathbb{R}, f\in C^2, f'(0)<0, f''(x)<0$ for all $x\in\mathbb{R}$. Let $\alpha\in(0,1)$. Prove that, for all $x>0$, $$f(x)\le f(0) + \alpha x f'(0).$$
This is what I came up with:
$$\frac{f(x)-f(0)}{x-0} = f'(c),\ c\in (0,x)\implies\\f(x) = f(0) + xf'(c)$$
using the same argument for $f'(x):$
$$\frac{f'(x)-f'(0)}{x-0} = f''(c_2), \ c_2\in(0,x)\implies f'(x) = f'(0) + xf''(c_2)$$
for $x=c$ we have $f'(c) = f'(0) + cf''(c_2)$ for $c_2\in(0,c)$
So
$$f(x) = f(0) + xf'(c) = f(0) + x(f'(0) + cf''(c_2)) \implies \\f(x) = f(0) + xf'(0) + xcf''(c_2)\le f(0)+\alpha f'(0)x + f''(c_2)cx\le f(0) + \alpha x f'(0)$$
for $\alpha\in (0,1)$
Your proof is fine. This is another one (with strict inequality). Since $f''(x)<0$ it follows that $f'$ is strictly decreasing and therefore, for $x>0$, $$\frac{f(x)-f(0)}{x-0} = f'(c)<f'(0)\implies f(x)< f(0) + xf'(0)< f(0) + \alpha xf'(0)$$ for any $\alpha\in(0,1)$ where in the last step we used the fact that $f'(0)<0$.