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Trigonometric inequality on a triangle

Let ABC be a triangle. Prove that

    \[\frac{1}{\sin^2 A} + \frac{1}{\sin^2 B} + \frac{1}{\sin^2 C} \geq 4\]

Solution

The function \displaystyle f(x) = \frac{1}{\sin^2 x } is convex, thus:

    \begin{align*} \frac{1}{3} \sum f(A) &= \frac{1}{3} \sum \frac{1}{\sin^2 A}\\ &\geq f\left ( \frac{1}{3} \sum A \right ) \\ &= \frac{1}{\sin^2 \left ( \frac{1}{3} \sum A \right )} \\ &= \frac{1}{\sin^2 \frac{\pi}{3}} \\ &= \frac{4}{3} \end{align*}

The result follows. In fact, something a bit stronger holds. Let r denote the inradius , R the circumradius and s the semiperimeter. Then,

    \[\frac{1} {\sin ^2 A} + \frac{1} {\sin^2 B} + \frac{1} {\sin^2 C} \geq \frac{2R}{r}\]

Indeed,

    \begin{align*} \sum \frac{1}{\sin^2 A} &= 4R^2\sum \frac{1}{a^2} \\ &\geq 4R^2 \sum \frac{1}{ab}\\ &=\frac{4R^2 \left ( a+b+c \right )}{abc}\\ &= \frac{8R^2s}{abc} \\ &= 8R^2 \cdot \frac{{\rm E}}{r} \cdot \frac{1}{4R{\rm E}} \\ &= \frac{2R}{r} \end{align*}

in view of the known identities

(1)   \begin{equation*} E = rs \end{equation*}

(2)   \begin{equation*} 4R\mathrm{E} = abc \end{equation*}

(3)   \begin{equation*} R \geq 2 r \end{equation*}

 

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