# TrigonometryTrig Inequality

#### jacks

##### Well-known member
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

##### Well-known member
MHB Math Helper
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.

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