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题目(第二届“友谊杯”国际数学竞赛题)已知正数a,b,c且满足a+b+c=1。证明:a~2/(b+c)+b~2/(a+c)+c~2/(a+b)≥1/2成立。证法1:降次法。a~2/(b+c)+b~2/(a+c)+c~2/(a+b)=(a[1-(b+c)])/(b+c)+(b[1-(a+c)])/(a+c)+(x[1-(a+b)])/(a+b)=a/(b+c)+b/(a+c)+c/(a+b)-(a+b+c)≥3/2-1=1/2。证法2:配方法。
The subject (2nd International Friendship Cup International Mathematical Contest Contest) has known positive a, b, c and satisfies a + b + c = 1. Proof: a ~ 2 / (b + c) + b ~ 2 / (a + c) + c ~ 2 / (a + b) ≥ 1/2 holds. Law 1: reduction method. a ~ 2 / b + c + b ~ 2 / a + c + c ~ 2 / a + b = a 1 - b + c + b + c + b 1 - a + c + a x + 1- a + b = a / b + c + b / a + c) + c / (a + b) - (a + b + c) ≧ 3/2 - 1 = 1/2. Law 2: with the method.