$f$ is twice differentiable, $f + 2 f^{'} + f^{''} \geq 0$ , prove the following Let $ f : [0,1] \rightarrow R$. $f$ is twice diff.
 and $f(0) = f(1) = 0$
If
 $f + 2 f^{'} + f^{''} \ge 0$ , prove that $f\le 0$ in the domain.
Please don’t give complete solution, only hints.

Hint Let $g(x)=f(x)e^x$. Then 
$$g''=(f+2f'+f'')e^x \geq 0 \,.$$
That means that $g$ is....
How does this solve the problem? 
