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Question Number 193546 by York12 last updated on 16/Jun/23

i f a + b + c = 1

f i n d m a x i m u m a b + b c + c a

a, b, c a r e n o n n e g a t i v e i n t e g e r s

Answered by Frix last updated on 16/Jun/23

a = b = 0 \land c = 1 \land d = + \infty \Rightarrow \max is + \infty

Commented by York12 last updated on 16/Jun/23

y e a h s i r y o u a r e r i g h t

b u t I j u s t f o r g o t t o w r i t e t h a t c o n d i t i o n

Answered by Subhi last updated on 16/Jun/23

A n o t h e r s o l u t i o n

{(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + a c)

a^{2} + b^{2} + c^{2} ⩾ a b + b c + a c

{(a + b + c)}^{2} ⩾ 3 (a b + b c + a c)

a b + b c + a c ⩽ \frac{{(a + b + c)}^{2}}{3} = \frac{1}{3}

Commented by York12 last updated on 16/Jun/23

t h n x

Commented by York12 last updated on 17/Jun/23

\sqrt{\frac{x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} + \dots + x_{n}^{2}}{n}} ⩾ \frac{x_{1} + x_{2} + x_{3} + x_{4} + \dots + x_{n}}{n}

\Rightarrow (x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} + \dots + x_{n}^{2}) ⩾ \frac{{(x_{1} + x_{2} + x_{3} + x_{4} + \dots + x_{n})}^{2}}{n}

\Rightarrow n (x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} + \dots + x_{n}^{2}) ⩾ {(x_{1} + x_{2} + x_{3} + x_{4} + \dots + x_{n})}^{2}

\Rightarrow (n - 1) (x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} + \dots + x_{n}^{2}) ⩾ 2 \sum_{1 ⩽ i <} \sum_{j ⩽ n} x_{i} x_{j}

\Rightarrow \frac{(n - 1)}{2} (x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} + \dots + x_{n}^{2}) ⩾ \sum_{1 ⩽ i <} \sum_{j ⩽ n} x_{i} x_{j}

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