$cosalpha+cosbeta+cosgamma > 1$ [Galileiana 05]

[TomSawyer]

Siano $alpha, beta, gamma$ gli angoli interni di un triangolo.

Dimostrare che $cosalpha+cosbeta+cosgamma > 1$.

[karl]

Si ha:
$cosalpha +cosbeta+ cosgamma-1=cosalpha +cosbeta+ cosgamma+cos(alpha+beta+gamma)$=
=$2cos((alpha+beta)/2)cos((alpha-beta)/2)+2cos((alpha+beta)/2+gamma)cos((alpha+beta)/2)=$
=$2cos((alpha+beta)/2)[cos((alpha-beta)/2)+cos((alpha+beta)/2+pi-(alpha+beta) )]$=
=$2cos((alpha+beta)/2)[cos((alpha-beta)/2)-cos((alpha+beta)/2 )]$=
=$2sin((gamma)/2)[2sin((alpha)/2)sin((beta)/2)]=4sin((alpha)/2)sin((beta)/2)sin((gamma)/2)>0$
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