for a,b,c>0 prove that (ab+bc+ca)^2 ≥3(a+b+c)abc


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Question Number 17260 by Umar math last updated on 03/Jul/17

$f o r a, b, c > 0 p r o v e t h a t$ ${(a b + b c + c a)}^{2} ⩾ 3 (a + b + c) a b c$
Commented byprakash jain last updated on 03/Jul/17

$As expained in answer by 18 ? 1 .$ $The condition a, b, c > 0 is never used .$
	Answered by 18±1 last updated on 05/Jul/17

	$x = a b, y = b c, z = c a$ $\Leftrightarrow {(x + y + z)}^{2} ⩾ 3 (x y + y z + z x)$ $\Leftrightarrow x^{2} + y^{2} + z^{2} ⩾ x y + y z + z x$ $\Leftrightarrow {(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2} ⩾ 0$
	Commented by18±1 last updated on 03/Jul/17

	$s o r r y 3 (x y + y z + z x) n o t a, b, c$
	Commented byUmar math last updated on 03/Jul/17

	$t h a n k y o u s i r, b u t a m k i n d o f c o n f u s e d .$
	Commented by1234Hello last updated on 03/Jul/17

	$I am in doubt that the result is proved$ $or not ?!$
	Commented byprakash jain last updated on 03/Jul/17

	$Explaination of above answer$ ${(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2} ⩾ 0$ $\Rightarrow x^{2} - 2 x y + y^{2} + y^{2} - 2 y z + z^{2} + x^{2} - 2 x y ⩾ 0$ $\Rightarrow 2 (x^{2} + y^{2} + z^{2}) ⩾ 2 (x y + y z + z x)$ $\Rightarrow x^{2} + y^{2} + z^{2} ⩾ x y + y z + z x$ $\Rightarrow x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x ⩾ x y + y z$ $+ z x + 2 x y + 2 y z + 2 z x$ $\Rightarrow {(x + y + z)}^{2} ⩾ 3 (x y + y z + z x)$ $Now put x = a b, y = b c, z = c a$ ${(a b + b c + c a)}^{2} ⩾ 3 a b c (a + b + c)$
	Commented byUmar math last updated on 03/Jul/17

	$t h a n k s a b u n c h s i r, a m c l e a r k n o w$
	Commented byTinkutara last updated on 04/Jul/17

	$Thanks Sir!$
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