Let $f:[0, \pi] \rightarrow \mathbb{R}$ continuous and differentiable function in $[0, \pi]$ with $f(0) \geq 1 $ and $f'(x) \geq -1$ for all $x \in ]0, \pi]$. Show that $f(x) \geq 1 -x$ for all $x \in [0, \pi]$ with mean value theorem. Is this function monoton ?

So from the mean value theorem follows that a x exists such that $f'(x) =\frac{f(\pi) - f(0) }{ \pi - 0 }$. So $\frac{f(\pi) - f(0)}{ \pi} \geq -1$ $\Rightarrow $ $f(\pi) \geq -\pi + f(0)$ since $f(0) \geq 1$ it follows that $f(\pi) \geq -2.141....$ But now I don't know what to do.