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?
On
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 $
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 $\alpha = \beta = \gamma = \frac \pi3$