设f‘’(x)＜0,a,b＞0,证f(a+b)+f(0)＜f(a)+f(b)

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设f‘’(x)＜0,a,b＞0,证f(a+b)+f(0)＜f(a)+f(b)设f‘’(x)＜0,a,b＞0,证f(a+b)+f(0)＜f(a)+f(b)设f‘’(x)＜0,a,b＞0,证f(a+b)+

设f‘’(x)＜0,a,b＞0,证f(a+b)+f(0)＜f(a)+f(b)
设f‘’(x)＜0,a,b＞0,证f(a+b)+f(0)＜f(a)+f(b)

设f‘’(x)＜0,a,b＞0,证f(a+b)+f(0)＜f(a)+f(b)
f‘’(x)＜0
那么当x1 [f(a+b)-f(a)]/b
f(b)-f(0)>f(a+b)-f(a)
得到：f(a+b)+f(0)＜f(a)+f(b)