TITLE: Prove that $f(x)\leq 0$ for all $x\in [0;1]$
QUESTION [5 upvotes]: Let $f:[0;1]\to \mathbb{R}$ be a twice differentiable function on $(0;1)$ such that $f(0)=f(1)=0$ and $f''(x)+2f'(x)+f(x)\geq 0$ for all $x\in (a;b)$.  Prove that $f(x)\leq 0$ for all $x\in [0;1]$

I was thinking about considering other function: $g(x)=f(x)e^{x}$.
Then we have $$g(0)=g(1)=0; g'(x)=[f(x)+f'(x)]e^x; g''(x)=[f(x)+2f'(x)+f''(x)]e^x$$ It seems relate to the problem's condition. But I don't know how to continue then.
I need some help to clarify this problem. Thanks so much!

REPLY [3 votes]: $f''(x)+2f'(x)+f(x) \geq 0$
Let: $g(x) = f'(x)+f(x) \implies g'(x)+g(x) \geq0 \implies e^x(g'(x)+g(x)) \geq 0 $
$\implies \dfrac{d}{dx}\left(e^xg(x)\right) \geq0 \implies \dfrac{d}{dx}\left(e^x(f'(x)+f(x))\right) \geq 0 \implies \dfrac{d^2}{dx^2}\left(e^xf(x)\right) \geq 0$
This means $e^xf(x)$ is concave upward in $[0,1].$ Since $f(0) = f(1) =0$ and $e^x$ is always postive, the only way for $e^xf(x)$ to be concave up is if $f(x)$ is non positive.