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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

	$(2) i t h i n k \frac{3 b (c^{3} + a^{3})}{c + a} . . . . ?$ $(3) . . . (4) W e i g h t e d A M w o r c k s$
	Commented byloveineq last updated on 27/Jun/21

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