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Inequality involving area of a triangle

Show that in any triangle ABC with area \mathcal{A} the following holds:

    \[\frac{1}{2} \left ( \frac{\sin A + \sin B + \sin C}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right ) \leq \mathcal{A}\]


Let s is the semiperimeter , R the circumradius and r the inradius. From the law of sines we find

(1)   \begin{equation*}  \sin A + \sin B + \sin C = \frac{s}{R} \end{equation*}

as well as

(2)   \begin{equation*} \mathcal{A} = rs \end{equation*}


    \[\frac{1}{2\mathcal{A}} \sum \sin A \leq \sum \frac{1}{a^2}\]

Substitute the preceding equalities into the last inequality  and simplifying we obtain

    \[\frac{1}{2rR} \leq \sum \frac{1}{a^2}\]


    \[\mathcal{A} = rs = \frac{abc}{4R}\]

the last inequality becomes

    \[\frac{a+b+c}{abc} = \frac{1}{bc} + \frac{1}{ca} + \frac{1}{ab} \leq \frac{1}{a^2}+\frac{1}{b^2} + \frac{1}{c^2}\]

which is true by the rearrangement inequality.


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