How to show two different definitions of $\alpha$-strongly convex are equivalent? In my optimization lecture notes I have the following definition:
Definition: Let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ be differentiable. The $f$ is strongly convex it $\exists$ a positive constant $\alpha > 0$ such that 
$$
 \langle \nabla f(y) - \nabla f(x),y-x  \rangle \geq \alpha||y-x||^2 \,\,\,\,\,,\,\,\,\forall x,y \in \mathbb{R}^n \tag{1}
$$
However, in the Online Linear Optimization via Smoothing on page 2, it says the function is $\beta$-strongly convex if 
$$
 f(y) -f(x) - \langle  \nabla f(x),y-x  \rangle \geq \frac{\beta}{2}||y-x||^2 \,\,\,\,\,,\,\,\,\forall x,y \in \mathbb{R}^n \tag{2}
$$
How can we show that (1) is equivalent to (2) with appropriate choice of $\beta$?

Let’s take $(1)$ and write $g_1=f-\frac{\alpha}{2}\|\cdot\|_2^2$. Then $(1)$ is equivalent to $\langle \nabla g_1(y)-\nabla g_1(x),\,y-x \rangle \geq 0$, ie $g_1$ convex. 
Do the same for $(2)$ with $g_2= f-\frac{\beta}{2}\|\cdot\|_2^2$. 

I will post the following derivation which is expressed in Mindlak's answer just to have better understanding:
From (1) we have:
$$
 \langle \nabla f(y)  -\nabla f(x),y-x  \rangle \geq \alpha||y-x||^2= 
\alpha \langle y - x,y-x  \rangle\,\,\,\,\,,\,\,\,\forall x,y \in \mathbb{R}^n 
$$
So,
$$
 \langle \nabla f(y) - \alpha y -(\nabla f(x)-\alpha x),y-x  \rangle \geq 0\,\,\,\,\,,\,\,\,\forall x,y \in \mathbb{R}^n \tag{3}
$$
(3) means $g_1(x)=f(x)-\frac{\alpha}{2} \|x\|_2^2$ is convex, because it is the monotonicity property of convex function. From 
$$
 g_1(s) \geq g_1(t) + \langle  \nabla g_1(t),s-t  \rangle  \,\,\,\,\,,\,\,\,\forall s,t \in \mathbb{R}^n 
$$
we have
$$
 f(y)-\frac{\alpha}{2} \|y\|_2^2 \geq f(x)-\frac{\alpha}{2} \|x\|_2^2 + \langle  \nabla f(x)-\alpha x,y-x  \rangle  \,\,\,\,\,,\,\,\,\forall y,x \in \mathbb{R}^n 
$$
Yields
$$
 f(y)-f(x)+ \langle  \nabla f(x),y-x  \rangle  \geq -\frac{\alpha}{2} \|x\|_2^2  +\frac{\alpha}{2} \|y\|_2^2  +\langle  -\alpha x,y-x \rangle =\frac{\alpha}{2} \|y-x\|_2^2
\\
\forall \,\,\,\,\,y,x \in \mathbb{R}^n 
$$
and $\alpha = \beta$