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It is known that a, B and C are the three sides of the triangle ABC. The comparison of the square of (a + B + C) is 2 (AB + BC + AC)
（a+b+c)²=a²+b²+c²+2ac+2ab+2bc=a²+b²+c²+2(ab+ac+bc)
Because a > 0, b > 0, C > 0
So we get a & # 178; + B & # 178; + C & # 178; > 0
From (a + B + C) &# 178; - 2 (AB + AC + BC) = A & # 178; + B & # 178; + C & # 178; > 0
So we get (a + B + C) &# 178; > 2 (AB + AC + BC)
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