Proving Gerretsen's Inequality Today in class we were shown Gerretsen's inequality: 
$$16Rr-5r^2\leq s^2 \leq 4R^2+4Rr+3r^2$$
Where $R$, $r$, and $s$ are the circumradius, in radius, and semiperimeter of a triangle. After some research, I found that this inequality was proven by showing the following:$$IH^2=4R^2+4Rr+3r^2-\frac{1}{4}\left(a+b+c\right)^2$$
$$9IG^2=\frac{1}{4}(a+b+c)^2-16Rr+5r^2$$
Where $IH$ is the distance between the incenter and the orthocenter, $IG$ is the distance between the incenter and the centroid, and $a$, $b$, and $c$ denote the sides of the triangle. How are these two equalities derived? Any help is very much appreciated. 

Your question is very interesting. Let me give my idea.
Firstly, we will compute $IG$. Note that we have $$\vec{GA} + \vec{GB} + \vec{GC} = \vec{0}.$$ Then, $$3\vec{IG} = \vec{IA} + \vec{IB} + \vec{IC}.$$
$$9IG^2 = \sum_{\circlearrowleft} IA^2 + 2\sum_{\circlearrowleft}\vec{IA}\vec{IB} $$
Recall that $IA^2 = r^2 + (s-a)^2$ (I don't go to this detail, but you can see it easily), and $2\vec{IA}\vec{IB} = IA^2 + IB^2 - AB^2$.
Then, one has: \begin{eqnarray}9IG^2 &=& 3\sum_{\circlearrowleft} IA^2 - \sum_{\circlearrowleft} a^2 = 9r^2 + 3[(s-a)^2 + (s-b)^2 + (s-c)^2] - (a^2+b^2+c^2) \\ &=& 9r^2 + 5s^2 - 4(ab+bc+ca). \ \ \ (1)\end{eqnarray}
Now, use the Heron's formula, we have $$S = sr = \sqrt{s(s-a)(s-b)(s-c)}$$.
So, $$sr^2 = s^3 - s^2(a+b+c) + s(ab+bc+ca) - abc = - s^3 + s(ab+bc+ca) -abc.$$
Note that $abc= 4RS = 4Rrs$. We get $$sr^2 = -s^3 + s(ab+bc+ca) -4Rrs$$ or $$ab+bc+ca = r^2+s^2+4Rr\ \ \ \ (2)$$
Substitute (2) to (1), we get $$9IG^2 = 9r^2 + 5s^2 - 4(r^2 + s^2 + 4Rr)  = 5r^2 + s^2 - 16Rr.$$
EDIT: Let me compute the $IH$. I use two lemmas.
Lemma 1 One has $$a\vec{IA} + b\vec{IB} + c\vec{IC} = \vec{0}.$$
Lemma 2 One has $$HA^2 = 4R^2 - a^2; HB^2 = 4R^2 - b^2; HC^2 = 4R^2-c^2.$$
I don't give the detail prove of two above lemmas, but you can see it like this: 


*

*lemma 1: the bisector of angle A divides the side $BC$ by ratio $b:c$, and so on with two other bisectors.

*lemma 2: it is easy to use the Euler line, which implies that $HA$ is two times of the distance from the circumcenter to the side $BC$.


Using lemma 1, we have $$2s \vec{HI} = a\vec{HA} + b\vec{HB} + c\vec{HC}.$$
Then, \begin{eqnarray}4s^2 HI^2 &=& \sum_{\circlearrowleft} a^2HA^2 + 2\sum_{\circlearrowleft} ab \vec{HA}\vec{HB}\\
&=& \sum_{\circlearrowleft} a^2(4R^2 - a^2) + \sum_{\circlearrowleft} ab (HA^2 + HB^2 -AB^2)\ \ \ \ \ \mbox{(using the lemma 2)}\\
&=& 4R^2(a^2+b^2+c^2) - (a^4+b^4+c^4) + \sum_{\circlearrowleft} ab (8R^2 - a^2-b^2-c^2) \\
&=& 4R^2(a^2+b^2+c^2 + 2ab +2bc +2ca) - (a^4+b^4+c^4) - (ab+bc+ca)(a^2+b^2+c^2) \\
&=& 16R^2s^2 - (a^4+b^4+c^4) - (ab+bc+ca)(a^2+b^2+c^2).\end{eqnarray}
I leave the part $(a^4+b^4+c^4) + (ab+bc+ca)(a^2+b^2+c^2)$ for you, because of the simple but long computation.
I give the answer, $$4s^2IH^2 = 16R^2s^2 + 16Rrs^2 + 12r^2s^2 - 4s^4.$$
Then, you get the equality.
