# Two-Triangle Inequality II

### Proof

We shall prove only the left inequality, as the right one is obtained from that by swapping $A\;$ with $A',\;$ $a\;$ with $a',\;$ etc.

Observe that $\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}\le 2\cos\frac{C}{2},\;$ implying that

$\displaystyle \sin A+\sin B+\sin C\le\cos\frac{A}{2}+\cos\frac{C}{2}+\cos\frac{C}{2}$

It follows by the Law of Sines that

\displaystyle\begin{align}\frac{a+b+c}{2R} &= \sin A+\sin B+\sin C\\ &\le \cos\frac{A}{2}+\cos\frac{C}{2}+\cos\frac{C}{2}. \end{align}

$\displaystyle \frac{3\sqrt{3}}{2}\ge\cos\frac{A'}{2}+\cos\frac{B'}{2}+\cos\frac{C'}{2}.$

Dividing one by another we obtain

$\displaystyle\frac{a+b+c}{3\sqrt{3}R}\le\frac{\displaystyle\cos\frac{A}{2}+\cos\frac{B}{2}+\cos\frac{C}{2}}{\displaystyle\cos\frac{A'}{2}+\cos\frac{B'}{2}+\cos\frac{C'}{2}}.$

### Acknowledgment

Dan Sitaru has kindly posted the above problem, with two proofs by Kevin Soto Palacios (Peru), at the CutTheKnotMath facebook page.

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