abc为正数,a+b>c 证a/(a+1)+b/(b+1)>c/(c+1)
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abc为正数,a+b>c 证a/(a+1)+b/(b+1)>c/(c+1)
abc为正数,a+b>c 证a/(a+1)+b/(b+1)>c/(c+1)
abc为正数,a+b>c 证a/(a+1)+b/(b+1)>c/(c+1)
构造函数:
f(x)=x/(x+1)
则;f(x)=x/(x+1)
=[(x+1)-1]/(x+1)
=1-[1/(x+1)]
由于1/(x+1)在 (0,正无穷)上单调递减
则:-[1/(x+1)]在 (0,正无穷)上单调递增
则:
f(x)=x/(x+1)在 (0,正无穷)上单调递增
由于:a+b>c 且abc为正数
则:f(a+b)>f(c)
即:(a+b)/[(a+b)+1]>c/(c+1)
[a/(a+b)+1]+[b/(a+b)+1]>c/(c+1)
又由于:
a/(a+1)>a/[(a+b)+1]
b/(b+1)>b/[(a+b)+1]
则:
a/(a+1)+b/(b+1)>a/[(a+b)+1]+b/[(a+b)=1]
又[a/(a+b)+1]+[b/(a+b)+1]>c/(c+1)
则:a/(a+1)+b/(b+1)>c/(c+1)
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