1问题的提出设△ABC的三边长与面积分别为a,b,c与△.则有a2+b2+c2≥43△(1)这是三角形不等式中众所周知的外森比克不等式,证明是很容易的.刘保乾在1996年针对这个不等式提出以下更强、更有意思的猜想(见文[1]):a2+b2+c2≥4△a2b2+b2c2+c2a2(2)本文给出不等式(2)的证明,并为几何不等式的证明构建一个数学模型.2预备知识在对不等式(2)的研究中,联想起了以下的Wolstenholme不等式:设x、y、z∈R,则对任意△A′B′C′成立着不 (1) Proposing that the trilateral length and area of ​​△ABC are a, b, c and △ respectively. Then there is a2+b2+c2≥43△ (1). This is a well-known insensibilism inequality in triangle inequality. It is very easy. Liu Baoqian proposed the following stronger and more interesting conjecture for this inequality in 1996 (see [1]): a2+b2+c2≥4Δa2b2+b2c2+c2a2 (2) In this paper, we give the inequality ( Proof of 2) and build a mathematical model for the proof of geometric inequalities. 2 Preliminary knowledge In the study of inequality (2), the following Wolstenholme inequality was recalled: Let x, y, and z∈R be any A’B’C’ holds true