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Trigonometric inequality

Prove that the following inequality holds in any triangle:

    \[\left ( \sin \frac{A}{2} + \sin \frac{B}{2} \right )^2 + \left ( \sin \frac{B}{2} + \sin \frac{C}{2} \right )^2 + \left ( \sin \frac{C}{2} + \sin \frac{A}{2} \right )^2 \leq 3\]

Solution

Let s denote the semiperimeter of the triangle. Using the cosine theorem we have that

    \[c^2=a^2+b^2-2ab \cos C = \left ( a-b \right )^2 + 4ab \sin^2 \frac{C}{2}\]

from which it follows that

(1)   \begin{equation*} \sin^2 \frac{C}{2} = \frac{\left ( s-a \right )\left ( s-b \right )}{ab}\end{equation*}

(2)   \begin{equation*} \sin^2 \frac{A}{2} = \frac{\left ( s-b \right )\left ( s-c \right )}{bc} \end{equation*}

(3)   \begin{equation*} \sin^2 \frac{B}{2} = \frac{\left ( s-a \right )\left ( s-c \right )}{ac} \end{equation*}

Thus, by Cauchy’s inequality we have:

    \begin{align*} \left ( \sin \frac{A}{2} + \sin \frac{B}{2} \right )^2 &= \frac{s-c}{c} \left ( \sqrt{s-b} \cdot \frac{1}{\sqrt{b}} +\sqrt{s-a} \cdot \frac{1}{\sqrt{a}} \right )^2 \\ &\leq \frac{s-c}{c} \cdot \left ( \frac{1}{a} + \frac{1}{b} \right )\cdot \left ( s-a+s-b \right ) \\ &= \frac{\left ( a+b-c \right )\left ( a+b \right )c}{2abc}\\ &= \frac{\left ( a^2+b^2 \right )c -c^2\left ( a+b \right ) +2abc}{2abc} \\ &= 1 + \frac{\left ( a^2+b^2 \right )c -c^2\left ( a+b \right )}{2abc} \end{align*}

Hence,

    \begin{align*} \sum \left ( \sin \frac{A}{2} + \sin \frac{B}{2} \right )^2 &\leq 3 + \sum \frac{\left ( a^2+b^2 \right )c -c^2\left ( a+b \right )}{2abc} \\ &=3 + \frac{1}{2abc} \left ( \sum \left ( a^2+b^2 \right )c - \sum c^2 \left ( a+b \right ) \right ) \\ &= 3 \end{align*}

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