Inequality and condition

**dhiab** · July 16th 2009, 01:50 AM

Prove : If

then :

**alexmahone** · July 16th 2009, 04:26 AM

$(a+b+c)^2 \geq 0$

$a^2+b^2+c^2+2ab+2bc+2ca \geq 0$

$1+2(ab+bc+ca) \geq 0$

$ab+bc+ca \geq -\frac{1}{2}$

$-(ab+bc+ca) \leq \frac{1}{2}$ ----------------- (1)

----------------------------------------------------------------------------

$(a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2-ab-bc-ca)$

$=2(1-ab-bc-ca)$

$\leq 2(1+\frac{1}{2})$

$\leq 2(\frac{3}{2})$

$\leq 3$

So, $(a-b)^2+(b-c)^2+(c-a)^2 \leq 3$

Hence proved.

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