djb86 wrote: If $\alpha,\beta,\gamma$ are the angles of an acute-angled triangle, prove that \[\sin \alpha + \sin \beta > \cos \alpha + \cos\beta + \cos\gamma.\] \[\sin \alpha + \sin \beta > \cos \alpha + \cos\beta + \cos\gamma?\] $90^\circ>\alpha >90^\circ-\beta>0^\circ\Rightarrow sin\alpha >cos\beta $,$sin\beta >cos\gamma,sin \gamma>cos\alpha $ $\Rightarrow \sin \alpha + \sin \beta + \sin\gamma> \cos \alpha + \cos\beta + \cos\gamma.$

djb86 wrote: If $\alpha,\beta,\gamma$ are the angles of an acute-angled triangle, prove that \[\sin \alpha + \sin \beta > \cos \alpha + \cos\beta + \cos\gamma.\]

djb86 wrote: If $\alpha,\beta,\gamma$ are the angles of an acute-angled triangle, prove that \[\sin \alpha + \sin \beta > \cos \alpha + \cos\beta + \cos\gamma.\] (Baltic Way 1991) $90^\circ>\alpha >90^\circ-\beta>0^\circ\Rightarrow sin\alpha >cos\beta $,$sin\beta >cos\alpha $, $\Rightarrow 0<1-sin\alpha <1-cos\beta ,0<1-sin\beta <1-cos\alpha $, $\Rightarrow (1-sin\alpha )(1-sin\beta )<(1-cos\alpha )(1-cos\beta )$, $\Rightarrow sin\alpha+sin\beta-sin\alpha sin\beta >cos\alpha+cos\beta -cos\alpha cos\beta $, $\Rightarrow sin\alpha+sin\beta>cos\alpha+cos\beta -cos(\alpha+\beta) $, $\Rightarrow \sin \alpha + \sin \beta > \cos \alpha + \cos\beta + \cos\gamma.$

djb86 wrote: If $\alpha,\beta,\gamma$ are the angles of an acute-angled triangle, prove that \[\sin \alpha + \sin \beta > \cos \alpha + \cos\beta + \cos\gamma.\] The inequality is equivalent to $\frac{s}{R} - \sin\gamma > 1 + \frac{r}{R}$ Since the triangle is acute : $s > 2R+r$ $\Leftrightarrow \frac{s}{R} > 2 + \frac{r}{R}$ and $ -\sin\gamma > -1$ , we get the desired inequality.

Why s＞2R＋r？

Takeya.O wrote: Why s＞2R＋r？ Because $\cos\alpha\cos\beta\cos\gamma=\frac{p^2-(2R+r)^2}{4R^2}$.

Thanks,arqady. But Why cosαcosβcosγ={p^2－（2R＋r）^2}/4R^2 ？

It's just easy algebra: $\cos\alpha=\frac{b^2+c^2-a^2}{2bc}$, $R=\frac{abc}{4S}$, $r=\frac{2S}{a+b+c}$, $p=\frac{a+b+c}{2}$ and $S=\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}$.

In acute triangle ABC , there is the inequality $ cosA +cosB +cosC > sinA +sinB -cosC $

djb86 wrote: If $\alpha,\beta,\gamma$ are the angles of an acute-angled triangle, prove that \[\sin \alpha + \sin \beta > \cos \alpha + \cos\beta + \cos\gamma.\] Michael Niland wrote: In acute triangle ABC , there is the inequality $ cosA +cosB +cosC > sinA +sinB -cosC $ If $\alpha,\beta,\gamma$ are the angles of an acute-angled triangle, prove that \[\sin \alpha + \sin \beta > \cos \alpha + \cos\beta + \cos\gamma >\sin \alpha + \sin \beta -cos\gamma .\]Interesting.