# 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 the Euler's inequality is given at Elias Lampakis, Am Math Monthly, 122 (9), November 2015, p 892. Now, continuing with this madness of deriving everything from the half-angle formulas, we offer 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.

Question: Is this proof of Euler's inequality known? Any reference?

Crossposted at MathSE.