(1/(x(x+1)))+(1/((x+1)(x+2)))+ (1/((x+2)(x+3)))≤(3/4)


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Question Number 79233 by jagoll last updated on 23/Jan/20

$\frac{1}{x (x + 1)} + \frac{1}{(x + 1) (x + 2)} +$ $\frac{1}{(x + 2) (x + 3)} ⩽ \frac{3}{4}$
Commented by john santu last updated on 23/Jan/20

$\frac{1}{x} - \frac{1}{x + 1} + \frac{1}{x + 1} - \frac{1}{x + 2}$ $+ \frac{1}{x + 2} - \frac{1}{x + 3} ⩽ \frac{3}{4}$ $\Rightarrow \frac{1}{x} - \frac{1}{x + 3} ⩽ \frac{3}{4}$ $\frac{4}{4 x (x + 3)} ⩽ \frac{x (x + 3)}{4 x (x + 3)}$ $\frac{x^{2} + 3 x - 4}{x (x + 3)} ⩾ 0$ $(i) x \neq 0 \land x \neq - 1 \land x \neq - 2 \land x \neq - 3$ $(i i) \frac{(x + 4) (x - 1)}{x (x + 3)} ⩾ 0$ $s o l u t i o n \Rightarrow (- \infty, - 4] \lor (- 3, 0) \lor$ $[1, \infty) \land x \neq - 1 \land x \neq - 2$
Commented by mathmax by abdo last updated on 23/Jan/20

$(e) \Leftrightarrow \frac{1}{x} - \frac{1}{x + 1} + \frac{1}{x + 1} - \frac{1}{x + 2} + \frac{1}{x + 2} - \frac{1}{x + 3} ⩽ \frac{3}{4} \Leftrightarrow$ $\frac{1}{x} - \frac{1}{x + 3} ⩽ \frac{3}{4} \Leftrightarrow \frac{3}{x (x + 3)} ⩽ \frac{3}{4} \Leftrightarrow \frac{1}{x (x + 3)} - \frac{1}{4} ⩽ 0$ $(b u t x \neq 0, - 1, - 2, - 3) \Rightarrow \frac{4 - x^{2} - 3 x}{x^{2} + 3 x} ⩽ 0 \Leftrightarrow \frac{x^{2} + 3 x - 4}{x^{2} + 3 x} ⩾ 0$ $x^{2} + 3 x - 4 = 0 \to Δ = 9 + 16 = 25 \Rightarrow x_{1} = \frac{- 3 + 5}{2} = 1$ $x_{2} = \frac{- 3 - 5}{2} = - 4$ $x^{2} + 3 x = 0 \to x^{(1)} = 0 a n d x^{(2)} = - 3$ $x - \infty - 4 - 3 0 1 + \infty$ $x^{2} + 3 x - 4 + - - - +$ $x^{2} + 3 x + - - + +$ $Q + + + - +$ $s o t h e s e t o f s o l u t i o n i s$ $S =] - \infty, - 4 [\cup] - 4, 3 [\cup] - 3, 0 [\cup [1, + \infty [$
Commented by mathmax by abdo last updated on 23/Jan/20

$a n d x \neq - 1 a n d - 2$
	Answered by mr W last updated on 23/Jan/20

	$\frac{1}{x} - \frac{1}{x + 1} + \frac{1}{x + 1} - \frac{1}{x + 2} + \frac{1}{x + 2} - \frac{1}{x + 3} ⩽ \frac{3}{4}$ $\frac{1}{x} - \frac{1}{x + 3} ⩽ \frac{3}{4}$ $\frac{1}{x (x + 3)} ⩽ \frac{1}{4}$ $c a s e 1 : x < - 3$ $x (x + 3) > 0$ $x^{2} + 3 x - 4 ⩾ 0$ $(x + 4) (x - 1) ⩾ 0$ $\Rightarrow x ⩽ - 4 o r x ⩾ 1 \Rightarrow x ⩽ - 4$ $c a s e 2 : - 3 < x < 0$ $x (x + 3) < 0$ $x (x + 3) ⩽ 4$ $(x + 4) (x - 1) ⩽ 0$ $\Rightarrow - 4 ⩽ x ⩽ 1 \Rightarrow - 3 < x < 0 (b u t \neq - 1, \neq - 2)$ $c a s e 3 : x > 0$ $x (x + 3) > 0$ $x^{2} + 3 x - 4 ⩾ 0$ $\Rightarrow x ⩽ - 4 o r x ⩾ 1 \Rightarrow x ⩾ 1$ $s u m m a r y o f s o l u t i o n :$ $x ⩽ - 4 o r - 3 < x < 0 (b u t \neq - 1, \neq - 2) o r x ⩾ 1$
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