ab+c+d+ba+c+d+cd+a+b+da+b+c=a2ab+ac+ad+b2ab+bc+bd+c2cd+ac+bc+d2ad+bd+cd≥(a+b+c+d)22ab+2bc+2cd+2da+2ac+2bd
Ta sẽ chứng minh
(a+b+c+d)22ab+2bc+2cd+2da+2ac+2bd≥43
⇔3(a+b+c+d)2≥8(ab+bc+cd+da+ac+bd)
⇔3(a2+b2+c2+d2)−2(ab+bc+cd+da+ac+bd)≥0
⇔(a−b)2+(b−c)2+(c−d)2+(d−a)2+(a−c)2+(b−d)2≥0, hiển nhiên đúng.