Consider $\mathbb{R}^n$ with the canonical scalar product $(\cdot,\cdot)$ and the associated euclidean norm. Let $f: \mathbb{R}^n \to \mathbb{R}$ of class $\mathcal{C}^1$ be $\omega$-convex, i.e. $\forall u,v \in \mathbb{R}^n$: $(\nabla f(u)- \nabla f(v), u-v) \geq \omega \|u-v\|^2$.
I have to show that $\forall x \in \mathbb{R}^n$ : $f(x) \geq \frac{\omega}{2} \|x\|^2 - \| \nabla f(0) \| \|x\| + f(0)$.
I started by introducing the function $g(t)=f(tx)$ for $t \in [0,1]$. Applying the Taylor expansion on g gave me: $g(1) = g(0) + \frac{1-0}{1!} g'(0)$
Then I've calculated the derivative of g, I've found that $g'(t) = \sum_{i=1}^{n} \frac{\partial f}{\partial x_i}(xt) \cdot x_i$. Thus $g'(0) = (\nabla f(0),x)$.
Finally I've found $f(x) = f(0) + (\nabla f(0), x)$ but I don't know how to proceed to get the wanted result.
Working with the function $g$ is the right way. We find
$$ f(x)=g(1)=g(0)+\int_0^1 g'(t)\,\mathrm{d}t=f(0)+\int_0^1 (\nabla f(tx),x)\,\mathrm{d}t. $$
On the other hand, by assumption,
$$ (\nabla f(tx)-\nabla f(0),tx)\geq \omega \|tx\|^2, $$
which gives
$$ (\nabla f(tx),x)\geq (\nabla f(0),x)+\omega t \|x\|^2\geq -\|\nabla f(0)\|\|x\|+\omega t \|x\|^2. $$
Now just insert that inequality in the above integral and be done.