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

$$\sqrt{\mathit{x}+4}-\sqrt{\mathit{x}-1}>\sqrt{4\mathit{x}+5}$$

Commented by Frix last updated on 14/Nov/22

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

Answered by Acem last updated on 14/Nov/22

$$f\left(x\right)=\sqrt{4x+5};x\in [-\frac{5}{4},+\mathrm{\infty }[$$$$c_1(-1.25,0),c_2(0,\sqrt{5})$$$$$$$$g\left(x\right)=\sqrt{x+4}-\sqrt{x-1};x\in [1,+\mathrm{\infty }[$$$$c_3(1,\sqrt{5}),$$$$\underset{x\to +\mathrm{\infty }}{\lim }g\left(x\right)=\underset{x\to +\mathrm{\infty }}{\lim }\frac{(\sqrt{x+4}-\sqrt{x-1})(\sqrt{x+4}+\sqrt{x-1})}{\sqrt{x+4}+\sqrt{x-1}}$$$$=\underset{x\to +\mathrm{\infty }}{\lim }\frac{5}{\sqrt{x+4}+\sqrt{x-1}}=0$$$$\Rightarrow forx\in [1,+\mathrm{\infty }[:g\left(x\right)\in [\sqrt{5},0[$$$$\mathrm{\&}f\left(x\right)\in [3,+\mathrm{\infty }[$$$$\Rightarrow \sqrt{4x+5}isalwaysbiggerthang\left(x\right)\mathrm{\forall }x\in [1,+\mathrm{\infty }[$$$$Thenthestatement$$$$\sqrt{x+4}-\sqrt{x-1}>\sqrt{4x+5}iswrongorx\in \mathrm{\varnothing }$$$$$$

Commented by Acem last updated on 13/Nov/22

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