设f(x)在[a,b]上二阶可导,且f"(x)﹥0,取xi∈[a,b](i=1,2,...,n)及ki﹥0,(i=1,2,...,n)且满足k1+k2+...+kn=1.证明: f(k1x1+k2x2+...+knxn)≤k1f(x1)+k2f(x2)+.

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问题 设f(x)在[a,b]上二阶可导,且f"(x)﹥0,取xi∈[a,b](i=1,2,...,n)及ki﹥0,(i=1,2,...,n)且满足k1+k2+...+kn=1.证明:
f(k1x1+k2x2+...+knxn)≤k1f(x1)+k2f(x2)+...+knf(xn).

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答案令x0=k1x1+k2x2+...+knxn,显然x0∈[a,b],因为f"(x)>0,所以f(x)≥f(x0)+f’(x0)(x-x0),分别取x=xi(i=1,2,...n),得 [*]

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