Question

设函数f(x)的定义域为(0,+∞),x1>0,x2>0,求证：若f(x)/x单调下降,则f(x1+ x2)≤f(x1)

Answer 1

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)

f(x1+x2)/(x1+x2) ≤ f(x2)/x2 => x2*f(x1+x2)/(x1+x2) ≤f(x2)

两式相加得, (x1+x2)*f(x1+x2)/(x1+x2) ≤ f(x1)+f(x2),

故f(x1+ x2)≤f(x1)+f(x2)