Научно–методически статии
SOME INEQUALITIES IN THE TRIANGLE
Резюме. The paper considers some inequalities in the triangle connecting the semi-perimeter, the circumradius and the inradius.
Ключови думи: triangle, semi-perimeter, circumradius, inradius, inequality
The following inequalities from (Bottema et al., 1969) (5.5, 5.8, 5.11 and 5.12) are considered in the paper:
(1) \[s^{2} \geq 3 r(4 R+r) \]
(2) \[ s^{2} \geq r(16 R-5 r) \]
(3) \[ s^{2} \geq 27 r^{2} \]
(4) \[ s^{2} \geq \tfrac{27}{2} R r \]
The inequalities (1) – (4) are proved in (Bottema et al., 1969), (Grozdev, 2005), (Grozdev, 2007) and we ommit the proofs here. Our goal is to demonstrate that the inequality (2) is the best one with respect to the others. It is valid the follwing chain:
\[ s^{2} \geq r(16 R-5 r) \geq \tfrac{27}{2} R r \geq 3 r(4 R+r) \geq 27 r^{2} . \]
We have \(r(16 R-5 r) \geq \tfrac{27}{2} R r \Leftrightarrow 2(16 R-5 r) \geq 27 R \Leftrightarrow 5 R \geq 10 r \Leftrightarrow R \geq 2 r\) and the last is the well-known Euler’s inequality, whose several proofs one could find in (Arslanagić, 2004), (Arslanagić, 2006), (Arslanagić, 2008) and (Arslanagić, 2009). In a similar way:
\[ \tfrac{27}{2} R r \geq 3 r(4 R+r) \Leftrightarrow R \geq 2 r \text { and } 3 r(4 R+r) \geq 27 r^{2} \Leftrightarrow R \geq 2 r . \]
Now, we put the question: “Does any inequality, which is better than the inequality (2), exist or not?“ The answer is affirmative; this is the inequality (10) from (Mitrinović et al., 1989), p. 248, for a non-obtuse triangle:
(5) \[ s^{2} \geq 2 R^{2}+8 R r+3 r^{2} \]
The proof is as follows:
\[ 2 R^{2}+8 R r+3 r^{2} \geq r(16 R-5 r) \Leftrightarrow 2 R^{2}-8 R r+8 r^{2} \geq 0 \Leftrightarrow R^{2}-4 R r+4 r^{2} \geq 0 \Leftrightarrow(R-2 r)^{2} \geq 0 . \] The last inequality is obvious.
Note that the equalities in (1), (2), (3), (4) and (5) hold true if and only if \(R=2 r\), i.e. for the equilateral triangle. We propose two other proofs of (5).
Proof 1. We will use the next well-known equalities:
(6) \[ a^{2}+b^{2}+c^{2}=2\left(s^{2}-r^{2}-4 R r\right) \]
(7) \[ 4 F=\left(b^{2}+c^{2}-a^{2}\right) \operatorname{tg} \alpha \]
where \(F\) is the area of \(\triangle A B C\). The equality (7) could be deduced very easily by using the formulae \(F=\tfrac{b c}{2} \sin \alpha\) and \(\cos \alpha=\tfrac{b^{2}+c^{2}-a^{2}}{2 b c}\). We have \(4 F=\left(a^{2}+c^{2}-b^{2}\right) \operatorname{tg} \beta\) and \(4 F=\left(a^{2}+b^{2}-c^{2}\right) \operatorname{tg} \gamma\). Apply also the well-known facts:
(8) \[ \sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma=\tfrac{2 F}{R^{2}}, \]
(9) \[ \cos \alpha+\cos \beta+\cos \gamma=1+\tfrac{r}{R} . \]
We have: \(a^{2}+b^{2}+c^{2}=\left(b^{2}+c^{2}-a^{2}\right)+\left(c^{2}+a^{2}-b^{2}\right)+\left(a^{2}+b^{2}-c^{2}\right)=\)
\[ \stackrel{(7)}{=} 4 F\left(\tfrac{1}{\operatorname{tg} \alpha}+\tfrac{1}{\operatorname{tg} \beta}+\tfrac{1}{\operatorname{tg} \gamma}\right)=4 F\left(\tfrac{\cos \alpha}{\sin \alpha}+\tfrac{\cos \beta}{\sin \beta}+\tfrac{\cos \gamma}{\sin \gamma}\right)=8 F\left(\tfrac{\cos ^{2} \alpha}{\sin 2 \alpha}+\tfrac{\cos ^{2} \beta}{\sin 2 \beta}+\tfrac{\cos ^{2} \gamma}{\sin 2 \gamma}\right) \]
and by the Cauchy-Bunyakovski-Schwarz inequality (CBS inequality) we get: \[ \begin{aligned} & a^{2}+b^{2}+c^{2} \geq 8 F \cdot \tfrac{(\cos \alpha+\cos \beta+\cos \gamma)^{2}}{\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma} \stackrel{(8),(9)}{\geq} 8 F \cdot \tfrac{\left(1+\tfrac{r}{R}\right)^{2}}{\tfrac{2 F}{R^{2}}} \\ & \Leftrightarrow 2\left(s^{2}-r^{2}-4 R r\right) \geq 4(R+r)^{2} \Leftrightarrow s^{2} \geq 2 R^{2}+8 R r+3 r^{2} \end{aligned} \] which ends the proof.
Proof 2. Now we will use (9) and (10) together with
(10) \[ a \cos \alpha+b \cos \beta+c \cos \gamma=\tfrac{2 F}{R} . \]
By the CBS inequality e get:
\[ (a \cos \alpha+b \cos \beta+c \cos \gamma)\left(\tfrac{\cos \alpha}{a}+\tfrac{\cos \beta}{b}+\tfrac{\cos \gamma}{c}\right) \geq(\cos \alpha+\cos \beta+\cos \gamma)^{2} \] and consequently \(\tfrac{2 F}{R}\left(\tfrac{\cos \alpha}{a}+\tfrac{\cos \beta}{b}+\tfrac{\cos \gamma}{c}\right) \geq\left(1+\tfrac{r}{R}\right)^{2}\). By thesinelaw \(a=2 R \sin \alpha, b=2 R \sin \beta\) and \(c=2 R \sin \gamma\). It follows that: \(\tfrac{2 F}{R} \cdot \tfrac{1}{2 R}(\cot g \alpha+\cot g \beta+\cot g \gamma) \geq \tfrac{(R+r)^{2}}{R^{2}}\), which is equivalent to
(11) \[ \cot g \alpha+\cot g \beta+\cot g \gamma \geq \tfrac{(R+r)^{2}}{F} \]
Additionally, we also have: \(\cot g \alpha+\cot g \beta+\cot g \gamma=\tfrac{\cos \alpha}{\sin a}+\tfrac{\cos \beta}{\sin \beta}+\tfrac{\cos \gamma}{\sin \gamma}=\tfrac{\cos \alpha}{\tfrac{a}{2 R}}+\tfrac{\cos \beta}{\tfrac{a b}{2 R}}+\tfrac{\cos \gamma}{\tfrac{c}{2 R}}=2 R\left(\tfrac{\cos \alpha}{a}+\tfrac{\cos \beta}{b}+\tfrac{\cos \gamma}{c}\right)=\)
\(=2 R\left(\tfrac{b^{2}+c^{2}-a^{2}}{2 a b c}+\tfrac{a^{2}+c^{2}-b^{2}}{2 a b c}+\tfrac{a^{2}+b^{2}-c^{2}}{2 a b c}\right)=\tfrac{R}{a b c}\left(a^{2}+b^{2}+c^{2}\right)\) and from here, because \(a b c=4 R F\), we get:
\[ \cot g+c \cot g \beta+\cot g \gamma=\tfrac{R}{4 R F} \cdot 2\left(s^{2}-r^{2}-4 R r\right)=\tfrac{s^{2}-r(4 R+r)}{2 F} \]
Finally, we get: \(\tfrac{s^{2}-r(4 R+r)}{2 F} \geq \tfrac{(R+r)^{2}}{F} \Leftrightarrow s^{2} \geq 2 R^{2}+8 R r+3 r^{2}\), which ends the proof.
REFERENCES
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Bottema, O. and al. (1969). Geometric Inequalities. Groningen: Wolters-Noordhoff Publishing.
Mitrinović, D. S., J. E. Pečarić & V. Volenec (1989). Recdent Advances in Geometric Inequalities. Dordrecht/Boston/London: Kluwer Academic Publishers.
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