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Question Number 40344 by behi83417@gmail.com last updated on 20/Jul/18

Commented by MJS last updated on 20/Jul/18

$maybe we can show p ⩾ \frac{3}{2}$
Commented by behi83417@gmail.com last updated on 21/Jul/18

$l e t : x = t g A, y = t g B, z = t g C$ $n o w : t g A + t g B + t g C = t g A . t g B . t g C \Rightarrow$ $A + B + C = 180$ $i t i s j u s t a n i d i a .$
	Answered by MJS last updated on 20/Jul/18

	${it}^{'} s not true$ $x = 1$ $y = 4$ $z = \frac{5}{3}$ $p = \frac{283571}{51714} > 4$
	Commented by behi83417@gmail.com last updated on 20/Jul/18

	$d e a r M J S! p l e a s e c o n s i d e r t h e f i r s t$ $c o n d i t i o n : x + y + z = x y z .$
	Commented by MJS last updated on 20/Jul/18

	$I did :$ $1 + 4 + \frac{5}{3} = \frac{20}{3}$ $1 \times 4 \times \frac{5}{3} = \frac{20}{3}$
	Commented by MrW3 last updated on 20/Jul/18

	$x + y + z = x y z$ $\Rightarrow z = \frac{x + y}{x y - 1}$ $p = . . . = F (x, y)$ $F (x, y) h a s n o m a x i m a .$ $p = F (1, 4) = 5.48 . . . > 4$ $p = F (1, 100) = 4900.5 . . . > 4$ $p = F (1, 1000) = 499000.5 . . . > 4$ $s t a t e m e n t t h a t p < q (= 4 o r w h a t e v e r)$ $i s n o t t r u e .$
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