m>0,abc均大于0证明a/(a+m)+b/(b+m)大于c/(c+m)
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![m>0,abc均大于0证明a/(a+m)+b/(b+m)大于c/(c+m)](/uploads/image/z/6897539-11-9.jpg?t=m%3E0%2Cabc%E5%9D%87%E5%A4%A7%E4%BA%8E0%E8%AF%81%E6%98%8Ea%2F%28a%2Bm%29%2Bb%2F%28b%2Bm%29%E5%A4%A7%E4%BA%8Ec%2F%28c%2Bm%29)
m>0,abc均大于0证明a/(a+m)+b/(b+m)大于c/(c+m)
m>0,abc均大于0证明a/(a+m)+b/(b+m)大于c/(c+m)
m>0,abc均大于0证明a/(a+m)+b/(b+m)大于c/(c+m)
方法1
a,b,c,且m为正数
所以(a+m) (b+m) (c+m)都是大于0
要证a/(a+m) +b/(b+m)>c/(c+m)
即要a(b+m)*(c+m)+b(a+m)*(c+m)>c(a+m)(b+m)
即abc+abm+acm+amm+abc+abm+bcm+bmm-abc-acm-bcm-cmm>0
即abm+amm+abc+abm+bmm-cmm>0
又因为a+b>c mm>0
所以amm+bmm>cmm
所以abm+amm+abc+abm+bmm-cmm>0
得证
方法2
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
所以分母[(a+m)(b+m)(c+m)]>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)