# Algebra in Cantelli-Cheybyshev Inequality Proof

I am confused by the following (possibly simple) algebra in the proof of the Cantelli-Cheybyshev inequality. I am following Rohatgi and Saleh (2015, Section 3.4 Lemma 1) where we plug in $$\phi(x)=(x+c)^2$$ into Markov's inequality to get:
$$$$P(Y \geq t) \leq P\big( \phi(Y) \geq (t+c)^2 \big) \leq \underbrace{\frac{{E}[(Y+c)^2]}{(t+c)^2}}_{RHS}$$$$ where $$EY=0, t\geq 1,$$ and $$c>0$$. To find the tightest bound, we minimize the RHS wrt to $$c$$. This gives us $$$$c^* = \frac{Var Y}{t}$$$$ The proof in Rohatgi and Saleh (2015) and elsewhere (e.g. Cantelli's inequality proof) says plugging in $$c^*$$ gives the desired expression: $$$$P(Y \geq t) \leq \frac{Var Y}{Var Y + t^2}$$$$ But the algebra in between doesn't seem trivial (at least to me): if I plug in $$c^*$$ into RHS, I don't see how to get the RHS of Cantelli's inequality even knowing that I can add or substract $$EY=0$$: $$$$\frac{{E}[(Y+c)^2]}{(t+c)^2}\bigg|_{c=c^*} = \frac{Var Y + (Var Y/t)^2}{t^2 + 2t(Var Y/t) + (Var Y/t)^2} = ...?$$$$

Any help would be greatly appreciated!

$$t^2 + 2t(Var Y/t) + (Var Y/t)^2=(t+Var Y/t)^2=\frac{1}{t²}(t²+Var Y)^2$$