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

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

$${x}\:\:\in\:\:\mathbb{R}^{+} \:\left({positive}\:\:{real}\:\:{numbers}\right) \\ $$$${Prove}\:\:{that}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} \:+\:\frac{\mathrm{2}}{{x}}\:\:\geqslant\:\:\mathrm{3} \\ $$

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

$${x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{2} \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{3}−\mathrm{1} \\ $$$${x}^{\mathrm{3}} −\mathrm{1}−\mathrm{3}\left({x}−\mathrm{1}\right) \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)−\mathrm{3}\left({x}−\mathrm{1}\right) \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}−\mathrm{3}\right) \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}−{x}−\mathrm{2}\right) \\ $$$$\left({x}−\mathrm{1}\right)\left\{{x}\left({x}+\mathrm{2}\right)−\mathrm{1}\left({x}+\mathrm{2}\right)\right\} \\ $$$$\left({x}+\mathrm{2}\right)\left({x}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\left({x}+\mathrm{2}\right)>\mathrm{0}\:\:{and}\:\left({x}−\mathrm{1}\right)^{\mathrm{2}} \geqslant\mathrm{0}\:\:{when}\:{x}=\mathrm{1} \\ $$$$ \\ $$$${so}\:\:\:{x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{2}\geqslant\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}+\frac{\mathrm{2}}{{x}}\geqslant\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{2}}{{x}}\geqslant\mathrm{3} \\ $$$$ \\ $$

Answered by salahahmed last updated on 26/Apr/19

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