An Inequality In Triangle That Involves the Four Basic Centers

Proof 1

We'll assume a few basic facts as known:

\displaystyle\begin{align} &\cos A+\cos B+\cos C= 1+\frac{r}{R}\\ &\csc\frac{A}{2}+\csc\frac{B}{2}+\csc\frac{C}{2}\ge 6\\ &R\ge 2r\\ &m_a+m_b+m_c\ge 9r\\ &AH=2R\cos A,\;BH=2R\cos B,\;CH=2R\cos C,\\ &AG=\frac{2}{3}m_a,\;BG=\frac{2}{3}m_b,\;CG=\frac{2}{3}m_c.\\ \end{align}

Applying the above,

\displaystyle\begin{align} &\sum_{cycl}AH = 2R\sum_{cycl}\cos A=2R+2r\ge 6r\\ &\sum_{cycl}2\cdot AI=2r\sum_{cycl}\csc\frac{A}{2}\ge 12r\\ &\sum_{cycl}3\cdot AO=3(3R)\ge 18r\\ &\sum_{cycl}4\cdot AG=4\cdot\frac{2}{3}\sum_{cycl}m_a\ge 24r. &\end{align}

Adding all up gives the desired inequality. The equality holds for equilateral triangles.

Proof 2

Let $AB=c,\;BC=a,\;CA=b.\;$ Let $T\;$ be the Fermat-Torichelli point and $TA=a',\;$ $TB=b',\;$ $TC=c'.\;$ Then

(1)

$\displaystyle \sum_{cycl}(AH+2\cdot AI+3\cdot AO+4\cdot AG)\ge 10(a'+b'+c').$

On the other hand,

\displaystyle\begin{align} (a'+b'+c')^2 &\ge 3(a'b'+b'c'+c'a')\\ &=3\frac{4}{\sqrt{3}}[\Delta ABC]\\ &=4\sqrt{3}r^2\cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2}\\ &=4\sqrt{3}r^2\left(\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}\right)\\ &\ge 4\sqrt{3}r^2\cdot 3\cot\frac{A+B+C}{6}\\ &= 4\sqrt{3}r^2\cdot 3\sqrt{3}\\ &=36r^2, \end{align}

where we used the convexity of cotangent and Jensen's inequality. Thus

(2)

$\displaystyle a'+b'+c'\ge 6r.$

Together with (1) this proves the required inequality.

Acknowledgment

The inequality from the Romanian Mathematical Magazine has been posted by Dan Sitaru at the CutTheKnotMath facebook page. Dan later added a proof (Proof 1) by Kevin Soto Palacios. Proof 2 is by Rozeta Atanasova.