x ∈ R^+ (positive real numbers) Prove that x^2 + (2/x) ≥ 3


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Question Number 58600 by naka3546 last updated on 26/Apr/19

$x \in R^{+} (p o s i t i v e r e a l n u m b e r s)$ $P r o v e t h a t$ $x^{2} + \frac{2}{x} ⩾ 3$
	Answered by tanmay last updated on 26/Apr/19
	$x^3 −3x+2 x^3 −3x+3−1 x^3 −1−3(x−1) (x−1)(x^2 +x+1)−3(x−1) (x−1)(x^2 +x+1−3) (x−1)(x^2 +2x−x−2) (x−1){x(x+2)−1(x+2)} (x+2)(x−1)^2 (x+2)>0 and (x−1)^2 ≥0 when x=1 so x^3 −3x+2≥0 x^2 −3+(2/x)≥0 x^2 +(2/x)≥3$
	$x^{3} - 3 x + 2$ $x^{3} - 3 x + 3 - 1$ $x^{3} - 1 - 3 (x - 1)$ $(x - 1) (x^{2} + x + 1) - 3 (x - 1)$ $(x - 1) (x^{2} + x + 1 - 3)$ $(x - 1) (x^{2} + 2 x - x - 2)$ $(x - 1) {x (x + 2) - 1 (x + 2)}$ $(x + 2) {(x - 1)}^{2}$ $(x + 2) > 0 a n d {(x - 1)}^{2} ⩾ 0 w h e n x = 1$ $s o x^{3} - 3 x + 2 ⩾ 0$ $x^{2} - 3 + \frac{2}{x} ⩾ 0$ $x^{2} + \frac{2}{x} ⩾ 3$
	Answered by salahahmed last updated on 26/Apr/19

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