Inequality

For all real numbers a; b; c prove the following chain inequality

3(a² + b² + c²) â‰¥ (a + b + c)² â‰¥ 3(ab + bc + ca):

1 Answers

2305

Shaswata Roy ·2013-04-09 02:26:43

\frac{1}{2}[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]\geq 0(Sum of squares is +ve)

â†’ a^{2}+b^{2}+c^{2}\geq ab+bc+ca

Therefore,

3(a^{2}+b^{2}+c^{2})\geq (a^{2}+b^{2}+c^{2})+2(ab+bc+ca)\geq 3(ab+bc+ca)

â†’ 3(a^{2}+b^{2}+c^{2})\geq (a+b+c)^{2} \geq 3(ab+bc+ca)

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