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1963年,一道经典的不等式题在莫斯科数学竞赛中应运而生,原题如下:设 a,b,c∈R+,求证:a/(b+c)+b/(c+a)+c/(a+b)≥3/2.①这个不等式的证法很多,下面笔者给出两个最简单的证明过程.证法1:要证原不等式成立,只须证 a/(b+c)+1+b/(c+a)+1+c/(a+b)+1≥9/2,即只须证[2(a+b+c)](1/(b+d)+1/(c+a)+1/(a+b))≥9,由柯西不等式易知上式显然成立,所以原不等式
In 1963, a classic inequality problem came into being in the Moscow Mathematical Contest. The original questions were as follows: Suppose a, b, c ∈ R +, verify: a / b + c + b / c + a + c / (a + b) ≥3 / 2. ① There are a lot of evidences for this inequality, and the following two give the simplest proof procedure: Law 1: To prove the original inequality, just a / (b + c) 1 + b + c + a + 1 + c / (a + b) + 1 ≧ 9/2, that is, 1 / (c + a) +1 / (a + b)) ≥ 9, Cauchy inequality is easy to know the above formula is true, so the original inequality