在a,b,c是△ABC的三边之长,且a+b+c=m的条件下,来证明关于α(a2+b2+c2)+β(ab+bc+ca)+γabc形式的不等式的问题,笔者对此一类的问题进行探究和总结,得到四个结论和两个定理,供大家参考.先给出此条件下的几个结论.结论1 a2+b2+c2=m2-2(ab+bc+ca).结论2由均值不等式可得m=a+b+c≥ In the condition that a, b, c are the lengths of three sides of △ABC, and a+b+c=m, it is proved that α(a2+b2+c2)+β(ab+bc+ca)+γabc forms The inequality of the problem, the author of this type of problem to explore and summarize, to get four conclusions and two theorems, for your reference. First give several conclusions under this condition. Conclusion 1 a2+b2+c2=m2 -2(ab+bc+ca). Conclusion 2 From the mean inequality m=a+b+c≥