MHB Proving an Inequality Involving Sines

[juantheron]

Prove that $\sin \frac{A}{2}+\sin\frac{B}{2}+\sin \frac{C}{2}\leq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$

 

[Sudharaka]

Re: TRig Inequality

Prove that $\sin \frac{A}{2}+\sin\frac{B}{2}+\sin \frac{C}{2}\leq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$

Hi jacks, :)

I hope \(A,\,B\mbox{ and } C\) are the internal angles of a triangle and \(a,\,b,\,c\) are the sides opposite to those angles respectively. In that case you can use the Law of sines.

\[\frac{a}{\sin A} \,=\, \frac{b}{\sin B} \,=\, \frac{c}{\sin C} \!\]

\begin{eqnarray}

\frac{b}{a+c}&=&\frac{1}{\frac{a}{b}+\frac{c}{b}}\\

&=&\frac{1}{\frac{\sin A}{\sin B}+\frac{\sin C}{\sin B}} \\

&=&\frac{\sin B}{\sin A+\sin C}\\

\end{eqnarray}

Similarly,

\[\frac{c}{a+b}=\frac{\sin C}{\sin A+\sin B}\mbox{ and }\frac{a}{b+c}=\frac{\sin A}{\sin B+\sin C}\]

\[\therefore\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{\sin B}{\sin A+\sin C}+\frac{\sin C}{\sin A+\sin B}+\frac{\sin A}{\sin B+\sin C}\]

Using the sum to product identities and simplification gives,

\[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{ \sin\frac{B}{2}}{\cos\left(\frac{A-C}{2}\right)}+\frac{\sin\frac{A}{2}}{\cos\left( \frac{B-C}{2}\right)}+\frac{\sin\frac{C}{2}}{\cos\left( \frac{A-B}{2}\right)}~~~~~~~~~~~~~(1)\]

Note that, \(\displaystyle\left|\cos\left( \frac{B-C}{2}\right)\right|\leq 1\Rightarrow \frac{1}{\left|\cos\left( \frac{B-C}{2}\right)\right|}\geq 1\Rightarrow \frac{\sin \frac{A}{2}}{\left|\cos\left( \frac{B-C}{2}\right)\right|}\geq \sin \frac{A}{2}\Rightarrow -\sin \frac{A}{2}\geq\frac{\sin\frac{A}{2}}{\cos\left( \frac{B-C}{2}\right)}\mbox{ or }\frac{\sin\frac{A}{2}}{\cos\left( \frac{B-C}{2}\right)}\geq \sin \frac{A}{2}~~~~~~~~~~~~~(2)\)

Similarly, \[-\sin \frac{B}{2}\geq\frac{\sin\frac{B}{2}}{\cos\left( \frac{A-C}{2}\right)}\mbox{ or }\frac{\sin\frac{B}{2}}{\cos\left( \frac{A-C}{2}\right)}\geq \sin \frac{B}{2}~~~~~~~~~~~~~(3)\]

\[-\sin \frac{C}{2}\geq\frac{\sin\frac{C}{2}}{\cos\left( \frac{A-B}{2}\right)}\mbox{ or }\frac{\sin\frac{C}{2}}{\cos\left( \frac{A-B}{2}\right)}\geq \sin \frac{C}{2}~~~~~~~~~~~~~(4)\]

By (1), (2), (3) and (4),

\[\sin \frac{A}{2}+\sin\frac{B}{2}+\sin \frac{C}{2}\leq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\]

Kind Regards,
Sudharaka.