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Thread: Angles of a triangle

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

    Prove that if $\displaystyle A,B,C$ are angles of a triangle , then :
    $\displaystyle

    \begin{array}{l}
    1)\cos A + cosB + \cos C \le \frac{3}{2} \\
    2)\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} \le \frac{1}{8} \\
    \end{array}

    $
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  2. #2
    MHF Contributor red_dog's Avatar
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    1) $\displaystyle 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}+\cos C\leq\frac{3}{2}$

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

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

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

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

    2) Is the same inequality because

    $\displaystyle \cos A+\cos B+\cos C=4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}+1$
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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 $\displaystyle A,B,C\in[0,\frac{\pi}{2}]$.

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