文[1]给出了以下不等式:若a,b,c为满足a+b+c=1的正数,t≥1,则(ta~2+b)/(b+c)+(tb~2+c)/(c+a)+(ta~2+a)/(a+b)≥(t+3)/2.(1)文[2]改进了(1)式中的t的取值范围,指出只要t≥1/4,(1)式就成立.本文将进一步得到t的最佳取值范围,即当t≥1/(16～2(1/2)2~(1/2)+13)时,(1)式成立.证明如 The following inequality is given in [1]: if a, b, and c are positive numbers that satisfy a+b+c=1, t≥1, then (ta~2+b)/(b+c)+(tb ~2+c)/(c+a)+(ta~2+a)/(a+b)≥(t+3)/2.(1) The article [2] improves t in (1) The range of values ​​indicated that as long as t ≥ 1/4, the formula (1) is established. This article will further obtain the optimal range of t, that is, when t ≥ 1/(16~2(1/2)2~( When 1/2)+13), (1) is established. Proof