1.利用一次函数证明不等式 由一次函数y=kx+b的图像可知,如果 f(m)>0,f(n)>0,则对一切x∈(m,n)均有 f(x)>0,反之,如果f(m)<0,f(n)<0,则对 一切x∈(m,n)均有f(x)<0,把这一性质称 为保号性,利用一次函数的保号性可以证明一 些不等式. 例1 设a,b,c都是绝对值小于1的实 数,求证:ab+bc+ca>-1 (*) 证明 ∵ab+bc+ca+1 1. Use the first-order function to prove that the inequality is known by the image of the first-order function y=kx+b. If f(m)>0 and f(n)>0, then we have f(x) for all x∈(m,n). > 0, conversely, if f (m) <0, f (n) <0, then for all x ∈ (m, n) there is f (x) <0, this property is called the sign, use The guarantee of the first function can prove some inequalities. Example 1 Let a, b, and c be real numbers whose absolute value is less than 1. Prove that: ab+bc+ca>-1 (*) Prove ∵ab+bc+ca+1