Minimization on both sides of inequalities

I was reading a post regarding strong convexity here and was going through the following proposition:

Strong convexity of a continuously differentiable function $$f$$ implies the following condition: $$\frac{1}{2} ||\nabla f(x)||^2 \geq\mu (f(x)-f(x^*) ,\qquad \forall x$$
Proof:
Taking minimization with respect to y on both sides of
$$f(y)\geq f(x)+\nabla f(x)^T (y-x) + \frac{\mu}{2} ||y-x||^2$$
yields $$f(x^*)\geq f(x)-\frac{1}{2\mu} ||\nabla f(x) ||^2$$
and rearrange it will give us the proof

I don't really get the part where we were asked to perform minimization on both sides of the inequalities. How am I supposed to do that?

The reason is: If $$g,h : A \to \mathbb R$$ are functions with $$g(y) \le h(y) \qquad \forall y \in A$$ then $$\inf_{y \in A} g(y) \le \inf_{y \in A} h(y).$$
To prove this, take a sequence $$y_n$$ with $$h(y_n) \to \inf_{y \in A} h(y)$$. Then, $$\inf_{y \in A} g(y) \le g(y_n) \le h(y_n).$$ Now, $$n \to \infty$$ gives the desired inequality.
$$f(y)≥f(x)+∇f(x)^T(y−x)+\frac{μ}{2}||y−x||^2 ~~~~~~~~~~~(1)$$
with respect to $$y$$, we get $$y_{min{LHS}} = x^*$$, and we minimize the right-hand size with respect to $$y$$ via taking the partial derivative and setting it equal to $$0$$ to get $$y_{min_{RHS}} = x - \frac{1}{\mu}\nabla f(x)$$ and if we plug these two back into $$(1)$$ and rearange, then we can recover your first inequality.