if a;b;c>0 then: ((9abc)/(a^3 + b^3 + c^3 )) + (a^2 /(bc)) + (b^2 /(ca)) + (c^2 /(ab)) ≥ 6


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Question Number 144083 by mathdanisur last updated on 21/Jun/21

$i f a; b; c > 0 t h e n :$ $\frac{9 a b c}{a^{3} + b^{3} + c^{3}} + \frac{a^{2}}{b c} + \frac{b^{2}}{c a} + \frac{c^{2}}{a b} ⩾ 6$
	Answered by mitica last updated on 21/Jun/21

	$\frac{9 a b c}{a^{3} + b^{3} + c^{3}} + \frac{a^{2}}{b c} + \frac{b^{2}}{a c} + \frac{c^{2}}{a b} = \frac{9 a b c}{a^{3} + b^{3} + c^{3}} + \frac{a^{3} + b^{3} + c^{3}}{a b c} \overset{a m - g m}{⩾}$ $2 \sqrt{\frac{9 a b c}{a^{3} + b^{3} + c^{3}} \cdot \frac{a^{3} + b^{3} + c^{3}}{a b c}} = 2 \cdot 3 = 6$
	Commented bymathdanisur last updated on 21/Jun/21

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