The Euler's inequality is an immediate consequence of Euler's identity in a triangle,


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.



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