If abc=1 and a,b,c>0 prove that (a/(b^2 (c+1)))+(b/(c^2 (a+1)))+(c/(a^2 (b+1))) ≥ (3/2)


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Question Number 142906 by liberty last updated on 07/Jun/21

$I f a b c = 1 a n d a, b, c > 0 p r o v e$ $t h a t \frac{a}{b^{2} (c + 1)} + \frac{b}{c^{2} (a + 1)} + \frac{c}{a^{2} (b + 1)} ⩾ \frac{3}{2}$
	Answered by Snail last updated on 07/Jun/21

	$L e t u s r e c a l l {T i t u}^{'} s L e m m a$ $Σ \frac{x_{i}^{2}}{y_{i}} ⩾ \frac{{(Σ x_{i})}^{2}}{Σ y_{i}}$ $D e n o t e g i v e i n e a u a l i t y b y E$ $A s a b c = 1 M u l t i p l y i n g t h e i n e q u a l i t y b o t h s i d e$ $b y a^{2} b^{2} c^{2} a n d u s i n g i t s v a l u d 1 w e c a t r a n s f o r m$ $i t a n d g e t . . .$ $\frac{a^{3} c^{2}}{c + 1} + \frac{b^{3} a^{2}}{a + 1} + \frac{c^{3} b^{2}}{b + 1} ⩾ \frac{3}{2}$ $n o w l e t x_{1} = a^{2} c^{2} x_{2} = b^{2} a^{2} x_{3} = c^{2} b^{2}$ $y_{1} = \frac{c + 1}{a} y_{2} = \frac{a + 1}{b} y_{3} = \frac{b + 1}{c}$ $E ⩾ \frac{{(a^{2} c^{2} + b^{2} c^{2} + c^{2} a^{2})}^{2}}{\frac{c + 1}{a} + \frac{b + 1}{c} + \frac{a + 1}{b}} = Z (L e t)$ $N o w u s i n g t h e f a c t t h a t {(a^{2} c^{2} + b^{2} c^{2} + a^{2} b^{2})}^{2} ⩾ 9$ $a n d b y d o i n g L C M o f d e n o m i n a t o r w e c a n g e t t h a t$ $Z = \frac{3}{2}$ $S o E ⩾ Z e q u a l i t y h o l d s w h e n a l l \frac{x_{i}}{y_{i}} a r e s a m e$ $P r o v e d . . . .$
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