a/(a+m)+b/(b+m)-c/(c+m)(相减通分)
=[a(b+m)(c+m)+b(a+m)(c+m)-c(a+m)(b+m)]/[(a+m)(b+m)(c+m)]
因为三角形ABC三边长是a ,b, c>0,且m为正数困闭蔽
所以分态铅母[(a+m)(b+m)(c+m)]>汪州0
又因为a(b+m)(c+m)+b(a+m)(c+m)-c(a+m)(b+m)
=abc+abm+acm+am^2+abc+bam+bcm+bm^2-abc-cam-cbm-cm^2
=abc+(abm+bam)+(am^2+bm^2-cm^2)
因为a+b>c(三角形两边之和大于第三边)
所以am^2+bm^2=(a+b)m^2>cm^2
所以(am^2+bm^2-cm^2)>0
abc+(abm+bam)>0
所以a/(a+m)+b/(b+m)>c/(c+m)
sin(2a+b)/sina-2cos(a+b)=sinb/sina ==>左边=[sin(2a+b)-2cos(a+b)sina]/sina =[sina*cos(a+b)+sin(a+b)*cosa-2cos(a+b)sina]/sina =[sin(a+b)*cosa-cos(a+b)sina]/sina =sin(a+b-a)/sina=sinb/sina=右边
17证仔丛明：左式=｛2ab+(a+b)m｝/｛ab+(a+b)m+m²｝>｛ab+(a+b)m｝/｛ab+(a+b)m+m²｝
=1-m²/念余樱｛ab+(a+b)m+m²｝>1-m²/｛(a+b)m+m²｝>1-m²/｛(cm+m²｝=1-m/(c+m)=c/(c+m)
所以原命题得证
18证明：左式=cos(α+β）+sin(α+β)cosα/sinα-2cos(α+β)=｛sin(α+β)cosα-cos(α+β)sinα｝毁羡/sinα
=sinβ/sinα 所以原命题得证