## Inequalities in Triangle Geometry Theorem 1
Let \( a \), \( b \), \( c \) be the lengths of the sides of \( \triangle ABC \), \( R \), \( r \), \( r_a \), \( r_b \), \( r_c \) its circumradius, inradius, and the radii of the excircles corresponding to \( a \), \( b \), and \( c \). Denote by \( p \) the semi-perimeter of \( \triangle ABC \) and by \( S \) its area. Let \( h_a \), \( h_b \), \( h_c \) be the lengths of the altitudes, \( m_a \), \( m_b \), \( m_c \) the lengths of the medians, and \( l_a \), \( l_b \), \( l_c \) the lengths of the segments of the angle bisectors that belong to the triangle.
Then the following inequalities hold:
\begin{eqnarray*}
9r&\leq h_a+h_b+h_c\leq l_a+l_b+l_c\leq \sqrt{r_ar_b}+\sqrt{r **\( 1^{\circ} \)**Starting from \( 2S_{ABC}=\frac13(ah_a+bh_b+ch_c) \) and applying the Chebyshev’s inequality we get \( r(a+b+c)=2S_{ABC}\leq \frac19(a+b+c)(h_a+h_b+h_c) \). This implies the first of the inequalities. The equality holds if and only if the triangle is equilateral.**\( 2^{\circ} \)**The second inequality is obvious from \( l_a\geq h_a \), \( l_b\geq h_b \), \( l_c\geq h_c \). The equality holds if and only if the triangle is equilateral.**\( 3^{\circ} \)**Here we will use the problem \ref{problem_bisector_length} to get: \[ l_a=\frac{2\sqrt{bc}}{b+c}\cdot\sqrt{p(p-a)}\leq \sqrt{p(p-a)}=\frac{S}{\sqrt{(p-b)(p-c)}}= \sqrt{\frac{ S}{p-b}\cdot\frac S{p-c}}=\sqrt{r c}. \] The last equality is obtained using Heron’s formula. Now summing up the analogous relations for \( l_b \) and \( l_c \) we get the third inequality.**\( 4^{\circ} \)**Using \( (x+y+z)^2\leq 3(x^2+y^2+z^2) \) (this follows from \( x^2+y^2+z^2\geq xy+yz+zx \)) we get \( \sqrt{r_ar_b}+\sqrt{r c}+\sqrt{r_cr_a}\leq \sqrt 3\sqrt{r_ar_b+r c+r_cr_a} \). Furthermore, \begin{eqnarray*}r_ar_b+r c+r_cr_a&=&S^2\cdot\left(\frac1{(p-a)(p-b)}+ \frac1{(p-b)(p-c)}+\frac1{(p-c)(p-a)}\right)\\ &=&S^2\cdot\frac{p-a+p-b+p-c}{(p-a)(p-b)(p-c)}=p^2. \end{eqnarray*} The fourth inequality follows directly from the previous calculation.**\( 5^{\circ} \)**From \( p^2=r_ar_b+r c+r_cr_a\leq \frac13(r_a+r_b+r_c)^2 \) we get \( p\sqrt 3\leq r_a+r_b+r_c \). This implies the last inequality. The last equality follows from the incircle-excircle theorem (parts (b) and (c)).
Problem 1 The first two inequalities are proved in the previous theorem. The third one will follow once we prove that \( m_a\leq l_a \). But this was done in Problem 4 of Introduction. Using the inequality between arithmetic and quadratic mean we get \( \frac{m_a+m_b+m_c}3\leq \sqrt{\frac{m_a^2+m_b^2+m_c^2}3}= \sqrt{\frac{3a^2+3b^2+3c^2}{12}}=\frac12\sqrt{a^2+b^2+c^2}\leq \frac{3R}2 \). Here we used Problem 4 from Important results in geometry. Problem 2 Prove that \[ 27r^2\leq h_a^2+h_b^2+h_c^2\leq l_a^2+l_b^2+l_c^2\leq p^2\leq m_a^2+m_b^2+m_c^2\leq \frac{27}4R^2.\] From the first inequality in Theorem 1 we conclude \( 9r\leq h_a+h_b+h_c\leq 3\sqrt{\frac{h_a^2+h_b^2+h_c^2}3} \) (the last inequality follows from the inequality between the arithmetic and quadratic mean). After squaring we get \( 27r^2\leq h_a^2+h_b^2+h_c^2 \) which proves the first inequality in the sequence. The second follows from \( h_a\leq l_a \). The third inequality follows from \( l_a^2=\frac{4bcp(p-a)}{(b+c)^2}\leq p(p-a) \) and the two analogs. Adding them up yields to \( l_a^2+l_b^2+l_c^2\leq p(p-a+p-b+p-c)=p^2 \). The next inequality follows from Problem 5 and the last one from Problem 4 in Important results in geometry. Problem 3 Prove that \[ r\leq \frac{\sqrt{\sqrt 3 S}}{3}\leq \frac{\sqrt 3}9p\leq \frac12R.\] The first inequality in the sequence is equivalent to \( 9r^2\leq \sqrt 3 S \). Writing \( S=pr \) transforms this into \( 9r \leq \sqrt3 p \) and this follows from the theorem \ref{triangle_in}. The second inequality is equivalent to \( \sqrt 3 S\leq \frac13 p^2 \) or, after writing \( S=pr \), \( \sqrt 3 r\leq \frac13 p \). This is again equivalent to \( \sqrt 3p\geq 9r \). In order to prove the last inequality let us start from \( \frac{a+b+c}3\leq \sqrt{\frac{a^2+b^2+c^2}3}\leq \sqrt{\frac{9R^2}3}=R\sqrt 3 \). This implies that \( p=\frac{a+b+c}2\leq \frac{3R\sqrt3}2 \) and \( \frac{\sqrt 3 p}9\leq \frac19\cdot\frac12 \cdot 3R\cdot 3=\frac R2 \). Problem 4 Assume that \( M \) is the point inside the triangle \( ABC \). Let \( r \) be the inradius of the triangle. Prove that \( MA+MB+MC\geq 6r \). When does the equality hold? We know that \( MA+MB+MC\geq CB^{\prime} \) where \( B^{\prime} \) is the image of \( B \) under the rotation around \( A \) for \( 60^{\circ} \). We know that \begin{eqnarray*}B^{\prime}C^2&=&a^2+b^2-2ab\cos (60^{\circ}+\alpha)\\&=& a^2+b^2-2ab\cos60^{\circ}\cos \alpha+2ab\sin 60^{\circ}\sin\alpha\\&=& a^2+b^2-ab\cos\alpha +2\sqrt3 S_{ABC}\\&=&\frac12(a^2+b^2+c^2)+2\sqrt 3S_{ABC}.\end{eqnarray*} From Problem 5 in Important results in geometry we get \( \frac12(a^2+b^2+c^2)=\frac12\cdot\frac43\cdot(m_a^2+m_b^2+m_c^2)\geq \frac23p^2 \). From Theorem 1 we know that \( p\sqrt 3\geq 9r \) which implies that \( \frac12(a^2+b^2+c^2)\geq 18p^2 \). We also have \( 2\sqrt 3 S\geq 9r^2\cdot 2=18r^2 \) (problem 3) and this finally gives \( B^{\prime}C^2\geq 36r^2 \) which is equivalent to the desired inequality. The equality holds if and only if the triangle is isosceles. |

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