FIVE NEW PROOFS OF ONE TRIGONOMETRIC INEQUALITY IN THE TRIANGLE
Резюме. The paper considers five proofs of the inequality \(\cos \alpha+\cos \beta+\cos \gamma \leq \tfrac{3}{2}\).
Ключови думи: trigonometric inequality, triangle, proof
Consider the following inequality:
(1) \[ \cos \alpha+\cos \beta+\cos \gamma \leq \tfrac{3}{2}, \]
where \(\alpha, \beta\) and \(\gamma\) are the interior angles of the triangle \(\triangle A B C\).
Twelve different proofs of this inequality are given in Arslanagić from 2008. We propose five new.
Proof 1. It follows by the cosine theorem, that: \[ \begin{gathered} \tfrac{b^{2}+c^{2}-a^{2}}{2 b c}+\tfrac{c^{2}+a^{2}-b^{2}}{2 a c}+\tfrac{a^{2}+b^{2}-c^{2}}{2 a b} \leq \tfrac{3}{2} \\ \Leftrightarrow a\left(b^{2}+c^{2}-a^{2}\right)+b\left(c^{2}+a^{2}-b^{2}\right)+c\left(a^{2}+b^{2}-c^{2}\right) \leq 3 a b c \\ \Leftrightarrow a^{2} b+a b^{2}+b^{2} c+b c^{2}+a^{2} c+a c^{2} \leq a^{3}+b^{3}+c^{3}+3 a b c, \end{gathered} \]
Thus, we come to the Schur’s inequality (for \(\lambda=1\) ). The equality holds in (1) iff \(a=b=c\) and \(\cos \alpha=\cos \beta=\cos \gamma=\tfrac{1}{2}\), respectively, i.e. \(\alpha=\beta=\gamma=\tfrac{\pi}{3}\) (equilateral triangle).
Proof 2. It is shown in Proof 1, that
\[ \cos \alpha+\cos \beta+\cos \gamma \leq \tfrac{3}{2} \Leftrightarrow a^{2} b+a b^{2}+b^{2} c+b c^{2}+a^{2} c+a c^{2} \leq a^{3}+b^{3}+c^{3}+3 a b c . \] On the other hand \(a^{2} b+a b^{2}+b^{2} c+b c^{2}+a^{2} c+a c^{2}=2 s\left(s^{2}+r^{2}-2 R r\right)\), \(a^{3}+b^{3}+c^{3}=2 s\left(s^{2}-3 r^{2}-6 R r\right)\) and \(a b c=4 R r s\), where \(R\) and \(r\) are the circum radius and the in-radius of \(\triangle A B C\), respectively, while \(s\) is the semi-perimeter. We have:
\[ \begin{gathered} \cos \alpha+\cos \beta+\cos \gamma \leq \tfrac{3}{2} \Leftrightarrow 2 s\left(s^{2}+r^{2}-2 R r\right) \leq 2 s\left(s^{2}-3 r^{2}-6 R r\right)+12 R r s /: 2 s \\ \Leftrightarrow s^{2}+r^{2}-2 R r \leq s^{2}-3 r^{2}-6 R r+6 R r \Leftrightarrow 2 R r \geq 4 r^{2} \Leftrightarrow R \geq 2 r . \end{gathered} \]
The last is the well-known Euler’s inequality and this ends the proof.
Proof 3. We have: \(\cos \alpha+\cos \beta+\cos \gamma=\tfrac{b^{2}+c^{2}-a^{2}}{2 b c}+\tfrac{c^{2}+a^{2}-b^{2}}{2 a c}+\tfrac{a^{2}+b^{2}-c^{2}}{2 a b}=\)
\[ \begin{aligned} & =\tfrac{a\left(b^{2}+c^{2}\right)-a^{3}+b\left(c^{2}+a^{2}\right)-b^{3}+c\left(a^{2}+b^{2}\right)-c^{3}}{2 a b c} \\ & =\tfrac{(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)-2\left(a^{3}+b^{3}+c^{3}\right)}{2 a b c} \end{aligned} \]
On the other hand:
\(a+b+c=2 s, a b c=4 R r s ; a^{2}+b^{2}+c^{2}=2\left(s^{2}-r^{2}-4 R r\right) ; a^{3}+b^{3}+c^{3}=2 s\left(s^{2}-3 r^{2}-6 R r\right)\).
Thus,
\[ \begin{gathered} \cos \alpha+\cos \beta+\cos \gamma=\tfrac{2 s \cdot 2\left(s^{2}-r^{2}-4 R r\right)-2 \cdot 2 s\left(s^{2}-3 r^{2}-6 R r\right)}{2 \cdot 4 R r s} \\ =\tfrac{s^{2}-r^{2}-4 R r-s^{2}+3 r^{2}+6 R r}{2 R r}=1+\tfrac{r}{R} \end{gathered} \]
It follows from the Euler’s inequality \(R \geq 2 r\), that \(\tfrac{r}{R} \leq \tfrac{1}{2}\), hence \(\cos \alpha+\cos \beta+\cos \gamma \leq \tfrac{3}{2}\).
Proof 4. For arbitrary real numbers \(\alpha, \beta \in(0, \pi)\) we have:
\[ \begin{gathered} (\cos \alpha+\cos \beta-1)^{2}+(\sin \alpha-\sin \beta)^{2} \geq 0 \\ \Leftrightarrow \cos ^{2} \alpha+\cos ^{2} \beta+1+2 \cos \alpha \cos \beta-2 \cos \alpha-2 \cos \beta+\sin ^{2} \alpha-2 \sin \alpha \sin \beta+\sin ^{2} \beta \geq 0 \\ \Leftrightarrow\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)+\left(\sin ^{2} \beta+\cos ^{2} \beta\right)+1+2(\cos \alpha \cos \beta-\sin \alpha \sin \beta-\cos \alpha-\cos \beta) \geq 0 \\ \Leftrightarrow 1+1+1+2 \cos (\alpha+\beta)-2 \cos \alpha-2 \cos \beta \geq 0 \\ \Leftrightarrow \cos \alpha+\cos \beta-\cos (\alpha+\beta) \leq \tfrac{3}{2} \end{gathered} \]
But \(\alpha+\beta+\gamma=\pi\), i.e. \(\alpha+\beta=\pi-\gamma, \cos (\alpha+\beta)=\cos (\pi-\gamma)=-\cos \gamma\), , hence \(\cos \alpha+\cos \beta+\cos \gamma \leq \tfrac{3}{2}\).
Proof 5. We use the following well-known expressions:
\[ \cos 2 x=\tfrac{1-\operatorname{tg}^{2} x}{1+\operatorname{tg}^{2} x} \quad \text { and } \quad \cos x=\tfrac{1-\operatorname{tg}^{2} \tfrac{x}{2}}{1+\operatorname{tg}^{2} \tfrac{x}{2}} \]
It follows that: \(\cos \alpha+\cos \beta+\cos \gamma \leq \tfrac{3}{2} \Leftrightarrow \tfrac{1-\operatorname{tg}^{2} \tfrac{\alpha}{2}}{1+\operatorname{tg}^{2} \tfrac{\alpha}{2}}+\tfrac{1-\operatorname{tg}^{2} \tfrac{\beta}{2}}{1+\operatorname{tg}^{2} \tfrac{\beta}{2}}+\tfrac{1-\operatorname{tg}^{2} \tfrac{\gamma}{2}}{1+\operatorname{tg}^{2} \tfrac{\gamma}{2}} \leq \tfrac{3}{2}\). Use the substitutions: \(u=\operatorname{tg} \tfrac{\alpha}{2}, v=\operatorname{tg} \tfrac{\beta}{2}, w=\operatorname{tg} \tfrac{\gamma}{2} ;(u, v, w \gt 0)\) and the equality \(\operatorname{tg} \tfrac{\alpha}{2} \operatorname{tg} \tfrac{\beta}{2}+\operatorname{tg} \tfrac{\beta}{2} \operatorname{tg} \tfrac{\gamma}{2}+\operatorname{tg} \tfrac{\gamma}{2} \operatorname{tg} \tfrac{\alpha}{2}=1\)
We obtain \(u v+v w+u w=1\) and the inequality (1) is transformed to:
(2) \[ \tfrac{1-u^{2}}{1+u^{2}}+\tfrac{1-v^{2}}{1+v^{2}}+\tfrac{1-w^{2}}{1+w^{2}} \leq \tfrac{3}{2}, \]
because \(l+u^{2}=u v+v w+u w+u^{2}=(u+v)(u+w), l+v^{2}=(v+w)(v+u)\) and \(l+w^{2}=(w+u)(w+v)\). Thus
(3) \[ \begin{gathered} \tfrac{1-u^{2}}{(u+v)(u+w)}+\tfrac{1-v^{2}}{(v+w)(v+u)}+\tfrac{1-w^{2}}{(w+u)(w+v)} \leq \tfrac{3}{2} \\ \Leftrightarrow \tfrac{\left(1-u^{2}\right)(v+w)+\left(1-v^{2}\right)(u+w)+\left(1-w^{2}\right)(u+v)}{(u+v)(v+w)(u+w)} \leq \tfrac{3}{2} \\ \tfrac{v+w-u^{2} v-u^{2} w+u+w-u v^{2}-v^{2} w+u+v-u w^{2}-v w^{2}}{(u+v)(v+w)(u+w)} \leq \tfrac{3}{2} \\ \Leftrightarrow \tfrac{2(u+v+w)-\left[u^{2}(v+w)+v^{2}(u+w)+w^{2}(u+v)\right]}{(u+v)(v+w)(u+w)} \leq \tfrac{3}{2} . \end{gathered} \]
Using the arithmetic-geometric mean inequality, we have:
(4) \[ l=u v+v w+u w \geq 3 \sqrt[3]{u^{2} v^{2} w^{2}} \Leftrightarrow \sqrt{3} \geq 9 u v w . \]
From the inequality \[ (x+y+z)^{2} \geq 3(x y+y z+z x)\left(\Leftrightarrow \tfrac{1}{2}\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right] \geq 0\right) \] we get
(5) \[ u+v+w \geq \sqrt{3(u v+v w+u w)}=\sqrt{3 \cdot 1}=\sqrt{3} \]
Further, \(0 \lt (u+v)(v+w)(u+w)=(u+v+w)(u v+v w+u w)-u v w=u+v+w-u v w\) and \(0 \lt u^{2}(v+w)+v^{2}(u+w)+w^{2}(u+v)=(u+v+w)(u v+v w+u w)-3 u v w=u+v+w-3 u v w\), hence:
\(\tfrac{2(u+v+w)-(u+v+w-3 u v w)}{u+v+w-u v w} \leq \tfrac{3}{2} \Leftrightarrow \tfrac{u+v+w+3 u v w}{u+v+w-u v w} \leq \tfrac{3}{2} \Leftrightarrow u+v+w \geq 9 u v w, \quad\) which ends the proof.
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