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Question Number 180569 by SAMIRA last updated on 13/Nov/22

(√(x+4)) − (√(x−1)) > (√(4x+5))

$$\sqrt{\boldsymbol{{x}}+\mathrm{4}}\:−\:\sqrt{\boldsymbol{{x}}−\mathrm{1}}\:>\:\sqrt{\mathrm{4}\boldsymbol{{x}}+\mathrm{5}} \\ $$

Commented by Frix last updated on 14/Nov/22

false. the inequality is defined for x≥1 but for x≥0: (√(x+4))<(√(4x+5)) ⇒ for x≥1 (√(x+4))<(√(4x+5))∧(√(x−1))≥0 ⇒ (√(x+4))−(√(x−1))<(√(4x+5))

$$\mathrm{false}. \\ $$$$\mathrm{the}\:\mathrm{inequality}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}\geqslant\mathrm{1} \\ $$$$\mathrm{but}\:\mathrm{for}\:{x}\geqslant\mathrm{0}:\:\sqrt{{x}+\mathrm{4}}<\sqrt{\mathrm{4}{x}+\mathrm{5}} \\ $$$$\Rightarrow \\ $$$$\mathrm{for}\:{x}\geqslant\mathrm{1}\:\sqrt{{x}+\mathrm{4}}<\sqrt{\mathrm{4}{x}+\mathrm{5}}\wedge\sqrt{{x}−\mathrm{1}}\geqslant\mathrm{0} \\ $$$$\Rightarrow\:\sqrt{{x}+\mathrm{4}}−\sqrt{{x}−\mathrm{1}}<\sqrt{\mathrm{4}{x}+\mathrm{5}} \\ $$

Answered by Acem last updated on 14/Nov/22

f(x)= (√(4x+5)) ; x∈ [−(5/4), +∞[ c_1 (−1.25, 0), c_2 (0, (√5)) g(x)= (√(x+4)) − (√(x−1)) ; x∈ [1, +∞[ c_3 (1, (√5)) , lim_(x→+∞) g(x)= lim_(x→+∞) ((((√(x+4)) − (√(x−1)))((√(x+4)) + (√(x−1))))/( (√(x+4)) + (√(x−1)))) = lim_(x→+∞) (5/( (√(x+4)) + (√(x−1)))) =0 ⇒ for x∈ [1, +∞[ : g(x) ∈ [(√5) , 0[ & f(x) ∈ [3, +∞[ ⇒ (√(4x+5)) is always bigger than g(x) ∀x∈ [1, +∞[ Then the statement (√(x+4)) − (√(x−1)) > (√(4x+5)) is wrong or x∈ ∅

$${f}\left({x}\right)=\:\sqrt{\mathrm{4}{x}+\mathrm{5}}\:\:;\:{x}\in\:\left[−\frac{\mathrm{5}}{\mathrm{4}},\:+\infty\left[\:\right.\right. \\ $$$${c}_{\mathrm{1}} \left(−\mathrm{1}.\mathrm{25},\:\mathrm{0}\right),\:{c}_{\mathrm{2}} \left(\mathrm{0},\:\sqrt{\mathrm{5}}\right) \\ $$$$ \\ $$$${g}\left({x}\right)=\:\sqrt{{x}+\mathrm{4}}\:−\:\sqrt{{x}−\mathrm{1}}\:;\:{x}\in\:\left[\mathrm{1},\:+\infty\left[\:\right.\right. \\ $$$$\:{c}_{\mathrm{3}} \left(\mathrm{1},\:\sqrt{\mathrm{5}}\right)\:, \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:{g}\left({x}\right)=\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:\frac{\left(\sqrt{{x}+\mathrm{4}}\:−\:\sqrt{{x}−\mathrm{1}}\right)\left(\sqrt{{x}+\mathrm{4}}\:+\:\sqrt{{x}−\mathrm{1}}\right)}{\:\sqrt{{x}+\mathrm{4}}\:+\:\sqrt{{x}−\mathrm{1}}} \\ $$$$\:\:=\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:\frac{\mathrm{5}}{\:\sqrt{{x}+\mathrm{4}}\:+\:\sqrt{{x}−\mathrm{1}}}\:=\mathrm{0} \\ $$$$\:\Rightarrow\:{for}\:{x}\in\:\left[\mathrm{1},\:+\infty\left[\:\::\:{g}\left({x}\right)\:\in\:\left[\sqrt{\mathrm{5}}\:,\:\mathrm{0}\left[\right.\right.\right.\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\&\:{f}\left({x}\right)\:\in\:\left[\mathrm{3},\:+\infty\left[\right.\right. \\ $$$$\:\Rightarrow\:\sqrt{\mathrm{4}{x}+\mathrm{5}}\:\:{is}\:{always}\:{bigger}\:{than}\:{g}\left({x}\right)\:\forall{x}\in\:\left[\mathrm{1},\:+\infty\left[\right.\right. \\ $$$${Then}\:{the}\:{statement}\: \\ $$$$\sqrt{{x}+\mathrm{4}}\:−\:\sqrt{{x}−\mathrm{1}}\:>\:\sqrt{\mathrm{4}{x}+\mathrm{5}}\:\:\:{is}\:{wrong}\:{or}\:{x}\in\:\emptyset \\ $$$$ \\ $$

Commented by Acem last updated on 13/Nov/22

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