if the sum of three positive real numbers is equal to their product, prove that at least one of the numbers is larger than 1.7.


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Question Number 206355 by mr W last updated on 12/Apr/24

$i f t h e s u m o f t h r e e p o s i t i v e r e a l$ $n u m b e r s i s e q u a l t o t h e i r p r o d u c t,$ $p r o v e t h a t a t l e a s t o n e o f t h e$ $n u m b e r s i s l a r g e r t h a n 1.7 .$
	Answered by A5T last updated on 12/Apr/24

	$a + b + c = a b c ⩾ 3 \sqrt[3]{a b c} \Rightarrow (a b c) ⩾ 3^{\frac{3}{2}}$ $O n e m u s t b e a t l e a s t \sqrt[3]{3^{\frac{3}{2}}} = \sqrt{3} \approx 1.732$
	Answered by mr W last updated on 12/Apr/24

	$s = a + b + c ⩾ 3 \sqrt[3]{a b c} = 3 \sqrt[3]{a + b + c} = 3 \sqrt[3]{s}$ $s^{\frac{2}{3}} ⩾ 3 \Rightarrow s ⩾ \sqrt{3^{3}} = \sqrt{27}$ $a t l e a s t o n e n u m b e r m u s t b e$ $⩾ \frac{s}{3} ⩾ \sqrt{3} > 1.7$
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