A differentiable $f$ function is stronglyconvex if and only if $$f(t x_1+ (1-t) x_2) \le t f(x_1) + (1-t) f(x_2) - \frac{t (1-t)m}{2}\Vert x_1-x_2\rVert^2,~t \in [0,1] \quad(1)$$
Other characterization is that is strongly convex if and only if
\begin{equation} g(x) = f(x)-\frac{m}{2}\lVert x \rVert^2~\text{is convex},~\forall x\quad(2) \end{equation}
So I want to prove that (1)$\Longleftrightarrow$(2). My try is to prove when $g$ is convex what happens, so applying the definition of convexity I obtain this \begin{align*} f(t x_1+ (1-t) x_2) -& \frac{m}{2} \parallel t x_1+ (1-t) x_2\parallel^2 \\&= g(t x_1+ (1-t) x_2) \\&\le t g(x_1) + (1-t) g(x_2) \\&= t f(x_1) - \frac{m}{2} t\parallel x_1\parallel^2 + (1-t) f(x_2) - \frac{m}{2} (1-t)\parallel x_2\parallel^2, ~\forall x_2,x_1, t\in [0,1]. \end{align*}
But I don't know how to continue.
Hint: take $ \frac{m}{2} \parallel t x_1+ (1-t) x_2\parallel^2 $ to RHS and expand $\parallel t x_1+ (1-t) x_2\parallel^2 $, as Euclidean norm and re-combine terms.
Expanding $\parallel t x_1+ (1-t) x_2\parallel^2 $ and combining with left out terms on right hand side $-t\parallel x_1 \parallel^2 - (1-t)\parallel x_2 \parallel^2$, we leave the terms $tf(x_1)$ and $(1-t)f(x_2)$ as they same as in (1).
$\parallel t x_1+ (1-t) x_2\parallel^2 -t\parallel x_1 \parallel^2 - (1-t)\parallel x_2 \parallel^2 $
$= t^2\parallel x_1\parallel^2 + (1-t)^2\parallel x_2\parallel^2 + t(1-t)(x_1^T x_2 + x_2^T x_1) -t\parallel x_1 \parallel^2 - (1-t)\parallel x_2 \parallel^2$
$= t(t-1)\parallel x_1\parallel^2 + t(t-1)\parallel x_2\parallel^2 + t(1-t)(x_1^T x_2 + x_2^T x_1) $
taking $t(1-t)$ common we have,
$= t(1-t)[ - \parallel x_1\parallel^2 - \parallel x_2\parallel^2 + x_1^T x_2 + x_2^T x_1]$
$= -t(1-t)\parallel x_1 - x_2\parallel^2$
combining with earlier $\frac{m}{2}$, we have
$ -\frac{m}{2} t(1-t)\parallel x_1 - x_2\parallel^2 $ we which want we want in (1). Now we do the same steps backwards we can go from (1) to (2) as well, so (1) is equivalent to (2).