文[1]介绍了这样一个不等式的证明:若xi0,i=1,2,3,且∑3i=1xi=1,则1+1x21+1+1x22+1+1x23≤1207.该文作者给出了一个较为复杂的证明.本人现给出一个简单的初等证明.证明先证明:对任意0x1有1+1x2≤2507(2-x)(*)去分母,整理得x3-2x2+x-247≤0,将左边因式分解得(x-13)2(x-43)≤0 The paper [1] introduces the proof of such an inequality: if xi0,i=1,2,3, and ∑3i=1xi=1, then 1+1x21+1+1x22+1+1x23≤1207. A more complex proof is given. I now give a simple elementary proof. The proof first proves that for any 0x1 there is 1+1x2 ≤ 2507(2-x)(*) to the denominator, and the result is x3-2x2+x- 247 ≤ 0, the left factorization (x-13) 2 (x-43) ≤ 0