Let x, y, z be positive numbers, and a=x(y−z)2, b=y(z−x)2, c=z(x−y)2. Prove that a2+b2+c2≥2(ab+bc+ca). (Posed by Tang Lihua)
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b+c−ac+a−ba+b−c=−(y+z)(z−x)(x−y),=−(z+x)(x−y)(y−z),=−(x+y)(y−z)(z−x), so (b+c−a)(c+a−b)(a+b−c)=−(y+z)(z+x)(x+y)[(y−z)(z−x)(x−y)]2≤0. We can get 2(ab+bc+ca)−(a2+b2+c2)=(a+b+c)(b+c−a)(c+a−b)(a+b−c)≤0. This means that a2+b2+c2≥2(ab+bc+ca).
Techniques
Linear and quadratic inequalitiesPolynomial operations