Proving $\csc\frac\alpha2+\csc\frac\beta2+\csc\frac\gamma2 \ge 6$, where $\alpha$, $\beta$, $\gamma$ are the angles of a triangle [closed]

If $\alpha$, $\beta$, $\gamma$ are angles of a triangle. prove that $\csc\frac\alpha2+\csc\frac\beta2+\csc\frac\gamma2 \ge 6$.

I started from $\alpha + \beta + \gamma = 180^{\circ}$ and then I tried to use some trigonometric identities, but with no success. Can someone tell me how to start?

closed as off-topic by Carl Mummert, user21820, Brian Borchers, Mostafa Ayaz, TaroccoesbroccoJul 28 '18 at 10:49

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Check the convexity of the function: $f(x) = \csc x = \frac {1}{\sin x}$ on the interval $(0,\frac \pi2)$ and you'll get it is a convex function. Then apply the Jensen's Inequality and you'll get:

$$\csc(\frac\alpha2)+\csc(\frac \beta2)+\csc(\frac \gamma2) \ge 3 \csc(\frac {\alpha + \beta + \gamma}{6}) = 3 \csc (\frac \pi6) = \frac{3}{\sin \frac\pi6} = \frac 3{\frac 12} = 6$$

Additionally from the Jensen's Inequality we can deduce that equality is reached if and only if $\alpha = \beta = \gamma = \frac \pi3$

• it makes sense now, thank you @Stefan4024 – Nawaf Feb 9 '15 at 9:47

If you're not familiar with Jensen's inequality here is an alternative approach.

First I'll prove that $\sin \frac{\alpha}{2} + \sin \frac{\beta}{2} +\sin \frac{\gamma}{2} \leq \frac{3}{2}$ where $\alpha, \beta , \gamma$ are interior angles in a triangle.

$\sin \frac{\alpha}{2} + \sin \frac{\beta}{2} +\sin \frac{\gamma}{2} = \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} + \cos\frac{\alpha + \beta }{2} = 2 \sin\frac{\alpha + \beta }{4}\cos \frac{\alpha - \beta }{4} +1-2 \sin^2 \frac{\alpha + \beta }{4}$

Now we have

$-2 \sin^2 \frac{\alpha + \beta }{4}+2 \sin\frac{\alpha + \beta }{4}\cos \frac{\alpha - \beta }{4}+1 \leq 2 \sin^2 \frac{\alpha + \beta }{4}+2 \sin\frac{\alpha + \beta }{4}+1 = \frac{3}{2} - 2 (\sin\frac{\alpha + \beta }{4}- \frac{1}{2})^2 \leq \frac{3}{2}$

Back to main problem: will use arithmetic-harmonic inequality (AH inequality)

$\csc(\frac\alpha2)+\csc(\frac \beta2)+\csc(\frac \gamma2)= \frac{1}{\sin\frac{\alpha}{2}}+\frac{1}{\sin\frac{\beta }{2}}+\frac{1}{\sin\frac{\gamma}{2}} \geq \frac{9}{\sin \frac{\alpha}{2} + \sin \frac{\beta}{2} +\sin \frac{\gamma}{2}} \geq \frac{9}{\frac{3}{2}} = 6$