高数,不等式证明,
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高数,不等式证明,
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![高数,不等式证明,](/uploads/image/z/19275630-6-0.jpg?t=%E9%AB%98%E6%95%B0%2C%E4%B8%8D%E7%AD%89%E5%BC%8F%E8%AF%81%E6%98%8E%2C%26nbsp%3B)
考虑g(x) = f(x+b)-f(x).
有g'(x) = f'(x+b)-f'(x).
由f"(x) > 0,f'(x)严格单调递增.
故g'(x) = f'(x+b)-f'(x) > 0,g(x)也严格单调递增.
于是g(a) > g(0),即f(a+b)-f(a) > f(b)-f(0) = f(b).
因此f(a)+f(b) < f(a+b).
有g'(x) = f'(x+b)-f'(x).
由f"(x) > 0,f'(x)严格单调递增.
故g'(x) = f'(x+b)-f'(x) > 0,g(x)也严格单调递增.
于是g(a) > g(0),即f(a+b)-f(a) > f(b)-f(0) = f(b).
因此f(a)+f(b) < f(a+b).