Minimum value of $\cos A+\cos B+\cos C$ in a triangle $ABC$

Question

I have used jensen's inequality but couldn't move on.

Answer 1

It is well-known that the sum of the distances of the circumcenter from the sides of a triangle equals the sum of the circumradius and the inradius, hence:

$$ R\left(\cos A+\cos B+\cos C\right) = R + r \tag{1}$$ and since $R\geq 2r$ by Euler's theorem, we have: $$ \cos A + \cos B+\cos C \leq \frac{3}{2} \tag{2} $$ with equality attained only by the equilateral triangle. Along the same lines, $$ \cos A+\cos B+\cos C\geq 1 \tag{3} $$ also follows. Equality is attained by a degenerate triangle with an angle equal to $\pi$.

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If you like to use Jensen's inequality, prove first that $\log\sin\frac{x}{2}$ is a concave function over $(0,\pi)$. Since $$ \cos A+\cos B+\cos C = 1+4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\tag{4}$$ the maximum of the LHS of $(4)$ is attained at $A=B=C$ and the minimum is attained on the boundary of the region $A,B,C\geq 0,A+B+C=\pi$.

Answer 2

$\cos A+\cos B+\cos C$

$=2\cos\dfrac{A+B}2\cos\dfrac{A-B}2+1-2\sin^2\dfrac C2$

$=2\sin\dfrac C2\cos\dfrac{A-B}2+1-2\sin\dfrac C2\cos\dfrac{A+B}2$ as $A+B=\pi-C,\dfrac{A+B}2=\dfrac\pi2-\dfrac C2$

$=1+2\sin\dfrac C2\left[\cos\dfrac{A-B}2-\cos\dfrac{A+B}2\right]$

$=1+4\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2$

Now $\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2$ can be made arbitrarily small, but$>0$ by setting $A,B\to0,C\to\pi$

Answer 3

$$ \begin{align} \sqrt[3]{\sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})} & \le \frac{\sin(\frac{A}{2}) + \sin(\frac{B}{2}) + \sin(\frac{C}{2}) }{3} \text{ (AM-GM inequality)}\\ & \le \sin \left( \frac{ \frac{A}{2} + \frac{B}{2} + \frac{C}{2}}{3} \right) \text{ (Jensen's inequality)}\\ & = \sin(\frac{\pi}{6})\\ & = \frac{1}{2} \end{align} $$

$$ 0 \lt \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2}) \le \frac{1}{8} $$

since

$$ \cos A+\cos B+\cos C=1+4\sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2}) $$

so we have: $$ 1 \lt \cos A+\cos B+\cos C \le \frac{3}{2} $$