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题已知x、y、z∈R+,x+y+z=1.求证:(1x2-x)(1y2-y)(1z2-z)≥(263)3.《中等数学》2006年第4期P48~49上刊登的解答较繁冗.下面,笔者给出一种贴近中学数学教学的简洁证明.证明因1-x=y+z≥2 yz,1+x+1x=1+x+19x+89x≥1+2x9x+89x=13(5+83x),对1+y+1y,1+z+1z有类似结论.故(1x2-x)(1y2-y
The problem is known as x, y, z∈R+, x+y+z=1. Proof: (1x2-x) (1y2-y) (1z2-z) ≥ (263) 3. “Middle Mathematics” No. 4 of 2006 The answer published in P48~49 is more verbose. In the following, the author gives a concise proof that is close to mathematics teaching in middle school. Prove that because 1-x=y+z≥2 yz, 1+x+1x=1+x+19x +89x≥1+2x9x+89x=13(5+83x), similar to 1+y+1y, 1+z+1z. Therefore (1x2-x)(1y2-y