Научно-методически статии
AN IMPROVEMENT OF THE GERRETSEN’S INEQUALITY FOR NON-OBTUSE TRIANGLES
Резюме. The inequality \(s^{2} \geq 2 R^{2}+8 R r+3 r^{2}\) is proved in the paper as a generalization of the Gerretsen’s en's inequality \(s^{2} \geq r(16 R-5 r)\) for (16 R - 5r ) for non-obtuse triangles.
Ключови думи: non-obtuse triangle, inequality, gеneralization
In (Bottema et al., 1969) the following inequality 5.14, p. 53 could be found:
(1) \[ 24 R r-12 r^{2} \leq a^{2}+b^{2}+c^{2} \leq 8 R^{2}+4 r^{2}, \]
where \(a, b, c\) a are the lenghts of the sides of the triangle \(\triangle A B C\) and \(R\) and \(r\) are the radius of the circumcircle and the radius of the incircleq respectively. This is the Gerretsen’sen's \(\mathbf{1}^{\mathbf{1}}\) inequality published in the Dutch journal Nieuw Tijdschr. Wisk. 41(1953), 1-7. By the well known equality \(a^{2}+b^{2}+c^{2}=2\left(s^{2}-r^{2}-4 R r\right)\) we rework (1) in the following form:
(2) \[ r(16 R-5 r) \leq s^{2} \leq 4 R^{2}+4 R r+3 r^{2}, \]
where \(s\) is the semi-perimeter of the triangle \(\triangle A B C\).
In the mathematical literature on inequalities, this inequality is named after Gerretsen and is known to be Gerretsen’s inequality.
In (Chirciu, 2015) the following series of inequalities 2.41., p. 22, are given: \[ \begin{aligned} & s^{2} \geq 16 R r-5 r^{2} \geq 15 R r-3 r^{2} \geq 14 R r-r^{2} \geq \tfrac{27}{2} R r \geq 13 R r+r^{2} \geq 12 R r+3 r^{2} \geq \\ & \geq 11 R r+5 r^{2} \geq 10 R r+7 r^{2} \geq 9 R r+9 r^{2} \geq 18 R r+11 r^{2} \geq 7 R r+13 r^{2} \geq 6 R r+15 r^{2} \geq \\ & \geq 5 R r+17 r^{2} \geq 4 R r+19 r^{2} \geq 3 R r+21 r^{2} \geq 2 R r+23 r^{2} \geq R r+25 r^{2} \geq 27 r^{2} . \end{aligned} \]
We will prove the inequality for non-obtuse triangle:
(3) \[ s^{2} \geq 2 R^{2}+8 R r+3 r^{2} \]
This inequality is stronger than the inequality \(s^{2} \geq r(16 R-5 r)\) from (2) because:
\[ \begin{aligned} & 2 R^{2}+8 R r+3 r^{2} \geq r(16 R-5 r) \\ & \Leftrightarrow 2 R^{2}-8 R r+8 r^{2} \geq 0 /: 2 \\ & \Leftrightarrow R^{2}-4 R r+4 r^{2} \geq 0 \\ & \Leftrightarrow(R-2 r)^{2} \geq 0 \end{aligned} \]
The last is obvious.
It follows, that equalities hold true in (2) and (3) iff \(a=b=c\), i. e. in the case of an equilateral triangle.
Proof of inequality (2):
By the Cauchy-Buniakowsky-Schwarz’s inequality we have: \[ \begin{gathered} \left((\sqrt{a \cos \alpha})^{2}+(\sqrt{b \cos \beta})^{2}+(\sqrt{c \cos \alpha})^{2}\right) \cdot\left(\left(\sqrt{\tfrac{\cos \alpha}{a}}\right)^{2}+\left(\sqrt{\tfrac{\cos \beta}{b}}\right)^{2}+\left(\sqrt{\tfrac{\cos \gamma}{c}}\right)^{2}\right) \geq(\cos \alpha+\cos \beta+\cos \gamma)^{2}, \text { i.e. } \\ (a \cos \alpha+b \cos \beta+c \cos \gamma)\left(\tfrac{\cos \alpha}{a}+\tfrac{\cos \beta}{b}+\tfrac{\cos \gamma}{c}\right) \geq(\cos \alpha+\cos \beta+\cos \gamma)^{2} \end{gathered} \] Further we apply the sine law:
\[ \begin{aligned} & a \cos \alpha+b \cos \beta+c \cos \gamma=2 R \sin \alpha \cos \alpha+2 R \sin \beta \cos \beta+2 R \sin \gamma \cos \gamma= \\ & =R(\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma)=4 R \sin \alpha \sin \beta \sin \gamma= \\ & \quad=\tfrac{2 R \sin \alpha \cdot 2 R \sin \beta}{R} \sin \gamma=\tfrac{2}{R} \cdot \tfrac{a b}{2} \sin \gamma=\tfrac{2 F}{R}, \end{aligned} \] where \(F\) is the area of \(\triangle A B C\).
Apply the sine law again:
\[ \begin{aligned} & \tfrac{\cos \alpha}{a}+\tfrac{\cos \beta}{b}+\tfrac{\cos \gamma}{c}=\tfrac{\cos \alpha}{2 R \sin \alpha}+\tfrac{\cos \beta}{2 R \sin \beta}+\tfrac{\cos \gamma}{2 R \sin \gamma}= \\ & =\tfrac{1}{2 R}(\operatorname{ctg} \alpha+\operatorname{ctg} \beta+\operatorname{ctg} \gamma) \text { and } \cos \alpha+\cos \beta+\cos \gamma=1+\tfrac{r}{R} . \end{aligned} \]
From (4) it follows, that:
\[ \begin{aligned} & \tfrac{2 F}{R} \cdot \tfrac{1}{2 R}(c \operatorname{tg} \alpha+c \operatorname{tg} \beta+c \operatorname{tg} \gamma) \geq\left(1+\tfrac{r}{R}\right)^{2} \\ & \Leftrightarrow c \operatorname{tg} \alpha+c \operatorname{tg} \beta+c \operatorname{tg} \gamma \geq \tfrac{(R+r)^{2}}{F} \end{aligned} \]
In (Mitrinović et al., 1989) the equality (69), p. 58 is given:
(6) \[ \operatorname{ctg} \alpha+\operatorname{ctg} \beta+\operatorname{ctg} \gamma=\tfrac{s^{2}-r^{2}-4 R r}{2 s r} . \]
From (5) and (6) we have:
\[ \begin{aligned} & \tfrac{s^{2}-r^{2}-4 R r}{2 s r} \geq \tfrac{(R \quad r)}{F} \\ & \Leftrightarrow \tfrac{s^{2}-r^{2}-4 R r}{2 F} \geq \tfrac{R^{2}+2 R r+r^{2}}{F} \\ & \Leftrightarrow s^{2} \geq 2 R^{2}+8 R r+3 r^{2} \end{aligned} \]
Which is the inequality (3), q.e.d.
Remark 1. The inequality (3) proposed by A. W. Walker in (Walker & Greening, 1972).
Remark 2. In (Mitrinović & Volenec, 1989), p. 58, proofs are given, that \(\operatorname{ctg} \alpha, \operatorname{ctg} \beta, \operatorname{ctg} \gamma\) are the roots of the equality
\[ 2 s r t^{3}-\left(s^{2}-r^{2}-4 R r\right) t^{2}+2 s r t+(2 R+r)^{2}-s^{2}=0 \]
It follows from here by the Vieta’s theorem that:
\[ \operatorname{ctg} \alpha+\operatorname{ctg} \beta+\operatorname{ctg} \gamma=\tfrac{s^{2}-r^{2}-4 R r}{2 s r} . \]
Remark 3. The following two equalities are well-known (Grozdev, 2005), (Grozdev, 2007):
\[ |H I|^{2}=4 R^{2}+4 R r+3 r^{2}-s^{2} \] and
\[ |G I|^{2}=\tfrac{1}{9}\left(s^{2}+5 r^{2}-16 R r\right), \] where \(H, I\) and \(G\) are the orthocentre, the incentre and the centroid of \(\triangle A B C\), respectively. The relations \(|H I| \geq 0\) and \(|G I|^{2} \geq 0\) give evidence of the Gerretsen's inequality (2).
NOTES
1. J. C. Gerretsen, (1907 – 1983), dutch mathematician
REFERENCES
Bottema, O., R. Ž. Djordjević, R. R. Janić, D. S. Mitrinović & P. M. Vasić (1969). Geometric Inequalities. Groningen (The Netherlands): WoltersNoordhoff Publisheng.
Chirciu, M. (2015). Inegalitati Geometrice, Pitesti (Romania): Editura Paralela 45.
Mitrinović, D. S., J. E. Pečarić & V. Volenec (1989). Recent Advances in Geometric Inequalities. Dordrecht/Boston/London: Kluwer Academic Publishers.
Walker, A. W. & M. G. Greening (1972). Problem E 2388, American Mathematical Monthly 70 (1972), 1135 and 80 (1973), 1142 – 1143.
Grozdev, S. (2005). Preparation for European Kangaroo. Sofia: UBM. (ISBN 954-8880-20-2), 220 pages (in Bulgarian).
Grozdev, S. (2007). For High Achievements in Mathematics. The Bulgarian Experience (Theory and Practice) . Sofia: ADE. (ISBN 978-954-921391-1), 295 pages.