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Question Number 152314 by mathdanisur last updated on 27/Aug/21
Answered by ghimisi last updated on 27/Aug/21
a2a2+ab+b2+c2c(a+b+c)⩾(a+c)2a2+b2+c2+ab+bc+acb2b2+cb+c2+a2a(a+b+c)⩾(a+b)2a2+b2+c2+ab+bc+acc2c2+ac+a2+b2b(a+b+c)⩾(b+c)2a2+b2+c2+ab+bc+ac⇒Σa2a2+ab+b2+1⩾2⇒Σa2a2+ab+b2⩾1
Commented by mathdanisur last updated on 27/Aug/21
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