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Math Help - Angles of a triangle

  1. #1
    Super Member dhiab's Avatar
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    Angles of a triangle

    Prove that if A,B,C are angles of a triangle , then :
     <br /> <br />
\begin{array}{l}<br />
1)\cos A + cosB + \cos C \le \frac{3}{2} \\ <br />
2)\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} \le \frac{1}{8} \\ <br />
\end{array}<br /> <br />
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  2. #2
    MHF Contributor red_dog's Avatar
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    1) 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}+\cos C\leq\frac{3}{2}

    2\sin\frac{C}{2}\cos\frac{A-B}{2}+1-2\sin^2\frac{C}{2}\leq\frac{3}{2}

    2\sin^2\frac{C}{2}-2\cos\frac{A-B}{2}\sin\frac{C}{2}+\frac{1}{2}\geq 0

    Let f(x)=2x^2-2\cos\frac{A-B}{2}x+\frac{1}{2}

    The discriminant is \Delta=4\left(\cos^2\frac{A-B}{2}-1\right)\leq 0\Rightarrow f(x)\geq 0, \ \forall x\in\mathbb{R}

    2) Is the same inequality because

    \cos A+\cos B+\cos C=4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}+1
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  3. #3
    Junior Member
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    A shorter solution for 1. It's enough to show that it's true for A,B,C\in[0,\frac{\pi}{2}].

    Now to the proof for this case. Notice that \cos x is a concave on [0,\frac{\pi}{2}] so by Jensens inequality:
    \frac{\cos A + \cos B + \cos C}{3} \leq \cos \left (\frac{A+B+C}{3} \right )=\cos \frac{\pi}{3} = \frac{1}{2} with equality iff A=B=C=\frac{\pi}{3}.
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