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Question Number 204640 by liuxinnan last updated on 24/Feb/24

$$f\left(x\right)=\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+a}}+\sqrt{\frac{ax}{ax+8}}$$$$a>0x>0$$$$prove1<f\left(x\right)<2$$

Answered by lepuissantcedricjunior last updated on 26/Feb/24

$$\mathit{x}>0\mathit{a}>0$$$$\mathit{x}<\mathit{x}+1;a<a+1=>0<\frac{\mathit{x}}{\mathit{x}+1}<\mathit{x};0<\frac{\mathit{a}}{1+\mathit{a}}<\mathit{a}$$$$=>0<\frac{1}{\sqrt{1+\mathit{x}}}<1;0<\frac{1}{\sqrt{1+a}}<1\left(1\right)$$$$0<a\mathbf{x}<\mathbf{ax}+8$$$$=>0<\frac{\mathbf{ax}}{\mathbf{ax}+8}<1=>0<\sqrt{\frac{a\mathit{x}}{\mathit{a}\mathit{x}+8}}<1\left(2\right)$$$$=>\left(1\right)+\left(2\right)$$$$=>0+0+0<\frac{1}{\sqrt{1+\mathbf{x}}}+\frac{1}{\sqrt{1+\mathbf{a}}}+\sqrt{\frac{\mathbf{ax}}{\mathbf{ax}+8}}<1+1+1$$$$=>0<\frac{1}{\sqrt{1+\mathbf{x}}}+\frac{1}{\sqrt{1+\mathbf{a}}}+\sqrt{\frac{\mathbf{ax}}{\mathbf{ax}+8}}<3$$$$=>0<1<\frac{1}{\sqrt{1+\mathbf{x}}}+\frac{1}{\sqrt{1+\mathbf{a}}}+\sqrt{\frac{\mathbf{ax}}{\mathbf{ax}+8}}<2<3$$$$1<\frac{1}{\sqrt{1+\mathbf{x}}}+\frac{1}{\sqrt{1+\mathbf{a}}}+\sqrt{\frac{\mathbf{ax}}{\mathbf{ax}+8}}<2$$$$=>1<\mathbf{f}\left(\mathbf{x}\right)<2$$$$.........profcedricjunior............$$$$$$$$$$

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