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Question Number 144634 by loveineq last updated on 27/Jun/21

Let a, b, c > 0 and (a + b) (b + c) = 4 . Prove that

(1) a^{2} + 2 b^{2} + c^{2} + \frac{2 b (c^{2} + a^{2})}{c + a} ⩾ 6

(2) a^{3} + 3 b^{3} + c^{3} + \frac{3 b (c^{3} + a^{3})}{c + a} ⩾ 8

(3) a^{4} + 4 b^{4} + c^{4} + \frac{4 b (c^{4} + a^{4})}{c + a} ⩾ 10

Answered by mindispower last updated on 27/Jun/21

c^{2} + a^{2} ⩾ \frac{1}{2} {(c + a)}^{2}

(1) \Leftrightarrow⩾ a^{2} + 2 b^{2} + c^{2} + b (c + a)

a^{2} + c^{2} ⩾ a c + \frac{1}{2} (a^{2} + c^{2})

⩾ (b^{2} + b c + a c + c b) + b^{2} + \frac{1}{2} (a^{2} + c^{2})

b^{2} + b c + a c + c b + \frac{a^{2} + c^{2}}{2} = t

(b + a) (b + c) + \frac{1}{4} (b^{2} + a^{2}) + \frac{1}{4} (b^{2} + c^{2}) + \frac{1}{2} b^{2} + \frac{1}{4} (a^{2} + c^{2})

⩾ 4 + \frac{1}{4} . 2 a b + \frac{1}{4} . 2 b c + \frac{1}{2} b^{2} + \frac{2 a c}{4} = 4 + \frac{1}{2} (b^{2} + a c + b c + a b)

= 4 + \frac{1}{2} (b + a) (b + c) = 6

\Leftrightarrow⩾ 6

Answered by mindispower last updated on 27/Jun/21

(2) i t h i n k \frac{3 b (c^{3} + a^{3})}{c + a} \dots . ?

(3) \dots (4) W e i g h t e d A M w o r c k s

Commented by loveineq last updated on 27/Jun/21

y e s, i s \frac{c^{3} + a^{3}}{c + a} .