Rewriting the definition of Strong Convexity

Question

Description

I've come across the following transition in a textbook of Convex Optimisation. I am struggling in figuring out how to transform in the equation below so that I'd appreciate if anyone hits me with any hint!

Problem

Definition of Strong Convexity

Suppose $x, y \in \mathbb{R}^n$, $f$ is said to be $\alpha$-convex if it satisfies the following;

$$ f(y) \leq f(x) + \nabla f(x)^T(y - x) + \frac{\alpha}{2} || y - x ||^2_2 $$

Then the transition of interest goes as

$$ f(y) \leq f(x) + \nabla f(x)^T(y - x) + \frac{\alpha}{2} || y - x ||^2_2 \\ \Longleftrightarrow \frac{\alpha}{2} || y - x +\frac{1}{\alpha} \nabla f(x) ||^2_2 \leq \frac{|| \nabla f(x) ||^2_2}{2 \alpha} - (f(x) - f(y)) $$

Answer 1

Silly me,,, I could solve it by myself by working backwards from the RHS to LHS as follows;

$$ \frac{\alpha}{2} || y - x +\frac{1}{\alpha} \nabla f(x) ||^2_2 \\ = \frac{\alpha}{2} \Big ( || y - x ||^2_2 + \frac{2}{\alpha} \langle + \nabla f(x) , y - x \rangle \frac{1}{\alpha^2} || \nabla f(x) ||^2_2 \Big ) \leq \frac{|| \nabla f(x) ||^2_2}{2 \alpha} - (f(x) - f(y)) \\ \Longleftrightarrow \frac{\alpha}{2} || y - x ||^2_2 + \langle \nabla f(x) , y - x \rangle \leq f(y) - f(x) $$