MHB Prove the trig inequality ∑α∈{A,B,C}1/[1+sin(α/2)]≥2

[lfdahl]

Prove, that for any triangle:

\[\sum_{\alpha \in \left \{ A,B,C \right \}}\frac{1}{1+\sin \frac{\alpha }{2}}\geq 2\]

 

[MarkFL]

My solution:

Spoiler

We have the objective function:

$$f(A.B,C)=\frac{1}{1+\sin\left(\dfrac{A}{2}\right)}+\frac{1}{1+\sin\left(\dfrac{B}{2}\right)}+\frac{1}{1+\sin\left(\dfrac{C}{2}\right)}$$

Subject to the constraint:

$$g(A,B,C)=A+B+C-\pi=0$$

We see by cyclic symmetry that the critical value occurs at:

$$A=B=C=\frac{\pi}{3}$$

And we find:

$$f\left(\frac{\pi}{3},\frac{\pi}{3},\frac{\pi}{3}\right)=\frac{3}{1+\sin\left(\dfrac{\pi}{6}\right)}=2$$

Using another point on the constraint, we find:

$$f\left(\frac{\pi}{6},\frac{\pi}{3},\frac{\pi}{2}\right)\approx2.04684845753400>2$$

Thus, we may assert:

$$f_{\min}=2$$

Shown as desired. :D

 

[lfdahl]

Well done, MarkFL! Thankyou for your participation!

Your response came before my ink dried ;)

May I also ask the forum participants for a solution without the use of cyclic symmetry?

 

[MarkFL]

Here's another solution:

Spoiler

Using the AM-GM inequality, we may state:

$$\frac{1}{3}\sum_{k=1}^3\left(\frac{1}{1+\sin\left(\dfrac{\theta_k}{2}\right)}\right)\ge\sqrt[3]{\prod_{k=1}^3\left(\frac{1}{1+\sin\left(\dfrac{\theta_k}{2}\right)}\right)}$$

We have that equality occurs for:

$$\theta_1=\theta_2=\theta_3=\frac{\pi}{3}$$

Hence:

$$\frac{1}{3}\sum_{k=1}^3\left(\frac{1}{1+\sin\left(\dfrac{\theta_k}{2}\right)}\right)\ge\frac{2}{3}$$

Or:

$$\sum_{k=1}^3\left(\frac{1}{1+\sin\left(\dfrac{\theta_k}{2}\right)}\right)\ge2$$

Shown as desired. :D

 

[lfdahl]

Thanks a lot, MarkFL, for a very fine solution!A third approach can be found here:

Spoiler

Note, that the function $\frac{1}{1+\sin x}$ is convex on $[0;\frac{\pi}{2}]$, and that the angles: $\frac{A}{2}$, $\frac{B}{2}$, $\frac{C}{2}$ all belong to the same interval.
Using Jensen´s inequality:

\[\frac{1}{1+\sin \frac{A}{2}}+\frac{1}{1+\sin \frac{B}{2}}+\frac{1}{1+\sin \frac{C}{2}} \geq \frac{3}{1+\sin \left (\frac{\frac{A}{2}+\frac{B}{2}+\frac{C}{2}}{3} \right ) }= \frac{3}{1+\sin 30^{\circ}}=2\]