a,b,c ,are nonnegative real numbers and: a+b+c=1 . show that: 0≤ ab+bc+ca−2abc ≤(7/(27)) .


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Question Number 54603 by behi83417@gmail.com last updated on 07/Feb/19

$a, b, c, a r e n o n n e g a t i v e r e a l n u m b e r s$ $a n d : a + b + c = 1 .$ $s h o w t h a t :$ $0 ⩽ ab + bc + ca - 2 abc ⩽ \frac{7}{27} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 08/Feb/19

	${(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + a c)$ $\frac{a + b + c}{3} ⩾ \sqrt[3]{a b c} \to \frac{1}{27} ⩾ a b c$ $\frac{a^{2} + b^{2} + c^{2}}{3} ⩾ \sqrt[3]{a^{2} b^{2} c^{2}}$ $\frac{a^{2} + b^{2} + c^{2}}{3} ⩾ {(a b c)}^{\frac{2}{3}} \approx {(\frac{1}{27})}^{\frac{2}{3}}$ $a^{2} + b^{2} + c^{2} \approx 3 \times \frac{1}{9}$ $s o 1 = \frac{1}{3} + 2 (a b + b c + a c)$ $a b + b c + a c \approx \frac{1}{3}$ $s o v a l u e o f a b + b c + a c - 2 a b c$ $= \frac{1}{3} - \frac{2}{27}$ $\frac{9 - 2}{27} = \frac{7}{27}$ $0 ⩽ a b + b c + a c - 2 a b c ⩽ \frac{7}{27}$
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