Q:

Show $f(x) \geq \sin(x)$ if $f(0)=0$, $f'(0)=1$ and $f''(x)+f(x)\geq 0$

Let $f:[0,\pi] \to \mathbb{R}$ be a twice differentiable function.
Show $f(x) \geq \sin(x) \forall x \in [0,\pi]$ if $f(0)=0$, $f'(0)=1$ and $f''(x)+f(x)\geq 0$ $\forall x \in [0,\pi]$.
I have tried making the taylor series.
Let $g(x) =f(x)-\sin(x)$. Then $g(x) =g(0)+g'(0) x +\frac{1}{2} g''(c) x^2=(f''(c)+\sin(c)) x^2/2 \geq (\sin(c)-f(c))x^2/2$
Thats all I can think of. Please help. Thank in advance.
EDIT: $\sin(x)$ is not unique since $e^x-1$ satisfies the condition too.
EDIT2: Conditions clarified.

A:

Let us denote $g=\left(f-\sin\right)$
  and assume that $g<0$
 . It is clear that $g\left(0\right)=0$, $g'\left(0\right)=0$
and that $g''\geq-g$
  with the hypothesis done on $f$
  and $f''$
 .
Now, let $x>0$
  be a non-negative element of $\left[0,\pi\right]$
 . By Taylor's formula with integral rest, one has$$0>g\left(x\right)=x^{2}\intop_{t=0}^{1}\left(1-t\right)g''\left(tx\right)dt\geq-x^{2}\intop_{t=0}^{1}\left(1-t\right)g\left(tx\right)dt.$$
 But $g\left(tx\right)\leq0$
  because $0\leq t\leq1$
 , so $-g\left(tx\right)\geq0$
 , whence $0>g\left(x\right)\geq0$,
 that is impossible. So the assumption $g<0$
  must be rejected, and then $g\geq0$
 .

