An Inequality in Triangle with the Circumradius, Inradius and Angle Bisectors

Problem

An Inequality in Triangle with the Circumradius, Inradius and Angle Bisectors

Proof

WLOG, $a\ge b\ge c.\,$ Clearly, $\ell_a=\min (\ell_a,\ell_b,\ell_c).\,$ We have

$\displaystyle\begin{align} R+r &= R+4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\\ &=R(\cos A+\cos B+\cos C) \end{align}$

and

$\displaystyle\begin{align} \ell_a &= \frac{2bc}{b+c}\cdot\cos\frac{A}{2}\\ &=\frac{4R\sin B\sin C}{\sin B+\sin C}\cdot\cos\frac{A}{2}=\frac{4R\sin B\sin C}{\displaystyle 2\sin\frac{B+C}{2}\cos\frac{B-C}{2}}\cdot\cos\frac{A}{2}\\ &=\frac{2R\sin B\sin C}{\displaystyle \cos\frac{B-C}{2}}. \end{align}$

We need to prove that $\displaystyle\cos A+\cos B+\cos C\ge\frac{2\sin B\sin C}{\displaystyle\cos\frac{B-C}{2}}.\,$ But

$\displaystyle\begin{align} 2\sin B\sin C &= \cos (B-C)-\cos (B+C)\\ &=\cos (B-C)+\cos (A) \end{align}$

and $\displaystyle \cos B +\cos C=2\sin\frac{A}{2}\cos\frac{B-C}{2}.\,$ Since $A\ge B\ge C,\,$ $A\ge\displaystyle\frac{\pi}{3},\,$ so that $\displaystyle\frac{A}{2}\ge\frac{\pi}{6},\,$ implying $\displaystyle\sin\frac{A}{2}\ge\frac{1}{2};\,$ also $B\,$ and $C\,$ are acute and, therefore $\displaystyle\frac{B-C}{2}\,$ is also acute, such that $\displaystyle\cos\frac{B-C}{2}\gt 0.\,$ We conclude that

$\displaystyle 2\sin \frac{A}{2}\cos\frac{B-C}{2}\ge\cos\frac{B-C}{2};$

hence, $\cos B+\cos C\ge\displaystyle\cos\frac{B-C}{2}.\,$ Thus, suffice it to show that

$\displaystyle \cos A+\cos\frac{B-C}{2}\ge\frac{\cos (B-C)+\cos A}{\displaystyle\cos\frac{B-C}{2}},$

or, equivalently,

$\displaystyle \cos^2\frac{B-C}{2}-\cos (B-C)\ge\left(1-\cos\frac{B-C}{2}\right)\cos A $

which is the same as

(*)

$\displaystyle \left(1-\cos\frac{B-C}{2}\right)\left(1+\cos\frac{B-C}{2}\right)\ge\left(1-\cos\frac{B-C}{2}\right)\cos A.$

Now, $\displaystyle \left(1-\cos\frac{B-C}{2}\right)\ge 0\,$ and $\displaystyle \left(1-\cos\frac{B-C}{2}\right)\gt 1\ge\frac{1}{2}\ge \cos A\,$ implying (*).

Acknowledgment

Leo Giugiuc has kindly communicated to me the problem by R. Satnoianu, with his solution. The problem has been previously published in the Romanian Mathematical Magazine by Dan Sitaru.

 

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