A convex function property

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.