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: your $g(x) = f(x) - \sin x$ satisfies $h = g'' + g \ge 0, g(0) = 0, g'(0) = 0.$  the integral representation gives $$g(x) = \int_0^x h(t)\sin (x-t) \, dt $$  that both $h$ and $\sin(x-t)$ are nonnegative for $0 \le x \le \pi$ implies $g(x) \ge 0$ on the same interval.
i can't see a way to extend this interval beyond $\pi.$
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 $g\geq0$
 .
A: $g(x)=f(x)-\sin x$ and $g(0)=0$
$g'(x)=f'(x)-\cos x$ and $g'(0)=0$
$g''(x)=f''(x)+\sin x$ and $g(x)+g''(x)\geq 0$ because $f(x)+f''(x)\geq 0$.
Let $x\in [0,\pi]$ be close to $0$ so, if we assume that $g(x)<0$ then due to continuity of $g$ at $x$ we have that $g(x)<0$ for all $x\in I_1$ , where $I_1$ is an interval and $0=\sup I_1$.
Then in $I_1$ we have that $g(x)<0\Rightarrow g''(x)>0\Rightarrow g'$ is increasing and thus $g'(x)\geq g'(0)=0\Rightarrow $ $g$ is increasing and thus $g(x)\geq g(0)=0$. Contradiction.
So $g(x)\geq 0 $ in $I_1$. Write $I_1=(0,a)$ and do the same for a $x$ close to $a$ where you know that there $g(x)\geq 0$.
Use induction to cover the interval $[0,\pi]$.
I could prove it more strictly the close to $0$ thing, but i don't know your  education level if you're at high school or something.
