a-b-c-are-nonnegative-real-numbers-and-a-b-c-1-show-that-0-ab-bc-ca-2abc-7-27-

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}