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#### sweatingbear

##### Member

- May 3, 2013

- 91

\(\displaystyle c \geqslant (a+b) \sin \left( \frac C2 \right)\)

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Let us stipulate \(\displaystyle 0^\circ < C < 180^\circ\) and of course \(\displaystyle a,b,c > 0\). Consequently, \(\displaystyle 0^\circ < \frac C2 \leqslant 90^\circ \implies 0 < \sin^2 \left( \frac C2 \right) \leqslant 1\) (we include \(\displaystyle 90^\circ\) for the angle \(\displaystyle \frac C2\) in order to account for right triangles).

Law of cosines yield

\(\displaystyle c^2 = a^2 + b^2 - 2ab \cos ( C )\)

Using \(\displaystyle \cos (C) \equiv 1 - 2\sin^2 \left( \frac C2 \right)\), we can write

\(\displaystyle c^2 = a^2 + b^2 - 2ab + 4ab\sin^2 \left( \frac C2 \right)\)

Now since \(\displaystyle 0 < \sin^2 \left( \frac C2 \right) \leqslant 1\), \(\displaystyle a^2 + b^2 - 2ab + 4ab\sin^2 \left( \frac C2 \right)\) will either have to equal \(\displaystyle a^2 + b^2 - 2ab + 4ab\) or be greater than it. Thus

\(\displaystyle c^2 \geqslant a^2 + b^2 - 2ab + 4ab = a^2 + 2ab + b^2 = (a+b)^2 \)

A similar argument can be made for \(\displaystyle (a+b)^2\) versus \(\displaystyle (a+b)^2 \sin^2 \left( \frac C2 \right)\): \(\displaystyle (a+b)^2\) will either have to equal or be greater than the latter expression, due to the values \(\displaystyle \sin^2 \left( \frac C2 \right)\) can assume. Therefore

\(\displaystyle c^2 \geqslant (a+b)^2 \geqslant (a+b)^2 \sin^2 \left( \frac C2 \right) \)

We can finally conclude

\(\displaystyle c \geqslant (a+b) \sin \left( \frac C2 \right)\)

Thoughts?