两式相加得, (x1+x2)*f(x1+x2)/(x1+x2) ≤ f(x1)+f(x2)。怎么加出来的 你用笔下 拍下 发邮件给我吧
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解:由f(x)的定义域为(0,+∞),x1>0,x2>0,
故x1+x2>x1,x1+x2>x2,由f(x)/x单调递减,
故f(x1+x2)/(x1+x2) ≤ f(x1)/x1 => x1*f(x1+x2)/(x1+x2) ≤ f(x1)①
f(x1+x2)/(x1+x2) ≤ f(x2)/x2 => x2*f(x1+x2)/(x1+x2) ≤f(x2)②
①+② 两式相加得, (x1+x2)*f(x1+x2)/(x1+x2) ≤ f(x1)+f(x2),
(这一步是把①和②中相同的因式 f(x1+x2)/(x1+x2))提取出来,然后把①剩下的x1和②剩下的x2相加)
故f(x1+ x2)≤f(x1)+f(x2)
(这一步是把式子等号左边的(x1+x2)*f(x1+x2)/(x1+x2)中分子,分母同除于(x1+x2))