Another proof of Euler inequality via the half-angle formulas

The Euler's inequality is an immediate consequence of Euler's identity in a triangle,
$$OI^2=R^2−2Rr.$$
An additional proof of Euler's inequality is given at Elias Lampakis, Am Math Monthly, 122 (9), November 2015, p 892. Now, continuing with this madness of proving everything from the half-angle formulas, we give an alternative proof of Euler's inequality.

Theorem (Euler): If $R$ and $r$ are the circumradius and, respectively, the inradius of a triangle, then
$$R\geq2r.\tag{1}$$

Proof. Let $a$, $b$ and $c$ be the sides of $\triangle{ABC}$. Denote $s$ its semiperimeter and $\angle{BAC}=\alpha$. It is easy to show that 

$$(s-a)\tan{\frac{\alpha}{2}}=r.\tag{2}$$

Also, it is well-known that

$$\frac{abc}{4\Delta}=R\qquad and \qquad\Delta=\frac{bc\sin{\alpha}}{2},\tag{3}$$

where $\Delta$ is the area of $\triangle{ABC}$. Substituting $(2)$ and $(3)$ in $(1)$ we have

$$\begin{aligned}\frac{abc}{4\Delta}&\geq(-a+b+c)\tan{\frac{\alpha}{2}}\\\frac{abc}{4bc\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}}&\geq(-a+b+c)\tan{\frac{\alpha}{2}}\\abc&\geq(-a+b+c)(4bc\sin^2{\frac{\alpha}{2}})\\abc&\geq(-a+b+c)(a-b+c)(a+b-c)\end{aligned}$$

This is Padoa's inequality, so the proof is complete.