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贵刊文[1]通过构造恒等式 a2b+c+b2c+a+c2a+b-a+b+c2 =(a+b+c)(ab+c+bc+a+ca+b-32)巧妙地证明了著名不等式(1)、(2)的等价性:命题1 (1963年莫斯科竞赛题)设a、b、c∈R+,求证: ab+c+b?
The article [1] skillfully proves the equivalence of the famous inequalities (1) and (2) by constructing the identity a2b+c+b2c+a+c2a+b-a+b+c2=(a+b+c)(ab+c+bc+a+ca+b-32): Proposition 1 (Moscow Competition of 1963) , b, c∈R+, verification: ab+c+b?