On a differential inequality Let $A>0$. If $f$ satisfies the differential inequality $f^{\prime\prime}(t)+f(t)≥A$, and $y$ is the solution to the ODE 
$y^{\prime\prime}(t)+y(t)=A$ with $y(0)=f(0)$ and $y^{\prime}(0)=f^{\prime}(0)$, then
\begin{equation}\label{GP}
 \begin{cases} 
f(t)\leq y(t) \quad \text{for all $t<0$}\\
f(t)\geq y(t) \quad \text{for all $t>0$}.
\end{cases} 
\end{equation}
Question: How can I prove this result?
 A: I don't think the result is true...
The general solution for the function $y(t)$ is
$$y(t)=C_1\sin(t) + C_2 \cos(t) +A$$
Let $$g(t) \equiv f(t)-y(t)$$
Then we have 
$$  g''(t)+g(t) \ge 0 $$
with initial conditions
$$ g(0)=g'(0)=0  $$
if $f(t) $ was a function satisfying
$$ f''(t)+f(t)=A+B $$
for some $B\ge 0$
Then its general solution would be
$$f(t)=C_3\sin(t) + C_4 \cos(t) +A+B$$
So that $$g(t) = (C_3-C_1)\sin(t) + (C_4 -C_2)\cos(t) +B $$
$$ g'(0)=0 \implies C_3-C_1=0  $$
$$ g(0)=0 \implies C_4-C_2=B  $$
So 
$$ g(t)=B \Big( 1+cos(t) \Big)$$
Notice that $g(t)\ge 0$ so that 
$$ f(t) \ge y(t)  $$
for all $t$.
A: Starting from $g''+g\ge 0$, which had been shown by @WW1 and @amsmath.
The roots of characteristic equation of corresponding equation are $\lambda_\pm \pm i$, thus the inequality can be rearranged by
\begin{equation}
g''-\lambda_- g' - \lambda_+ (g'-\lambda_- g)\ge 0
\end{equation}
Then by defining $G=g'-\lambda_- g$, we obtain
\begin{equation}
G'-\lambda_+ G\ge 0\to \frac{d}{d x}(e^{-x\lambda_+}G)\ge0
\end{equation}
which gives
\begin{equation}
G\ge G(0)e^{x\lambda_+},\qquad G(0)=g'(0)-\lambda_- g(0)
\end{equation}
namely, $G(0)$ depends on initial conditions. Now we shall solve
\begin{equation}
g'-\lambda_- g \ge G(0)e^{x\lambda_+}\to \frac{d}{d x}(e^{-x\lambda_-}g)\ge G(0)e^{x(\lambda_+-\lambda_-)}
\end{equation}
which gives
\begin{equation}
g\ge \frac{G(0)}{\lambda_+-\lambda_-} (e^{x\lambda_+}-e^{x \lambda_-})
=G(0) \sin(x).
\end{equation}
Therefore we arrive at
\begin{equation}
f\ge G(0) \sin(x)+y,\qquad y=A +[f(0) -A]\cos(x)+f'(0)\sin(x)
\end{equation}
This should be the final answer.
