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Question Number 149469 by mathdanisur last updated on 05/Aug/21

$$\mathrm{Compare}:$$$${1900}^{\frac{3}{4}}+{99}^{\frac{3}{4}}\mathbf{and}{1999}^{\frac{3}{4}}$$

Answered by mindispower last updated on 05/Aug/21

$$f\left(x\right)=x^{\frac{3}{4}}+{(1-x)}^{\frac{3}{4}},x\in [0,1[,f(1-x)=f\left(x\right)$$$$f^{\prime }\left(x\right)=\frac{3}{4}(\frac{1}{x^{\frac{1}{4}}}-\frac{1}{{(1-x)}^{\frac{1}{4}}})=\frac{3}{4}\left(\frac{{(1-x)}^{\frac{1}{4}}-x^{\frac{1}{4}}}{{\left(x(1-x)\right)}^{\frac{1}{4}}}\right)$$$$ifx⩽\frac{1}{2},f^{\prime }>0$$$$ifx\in ]\frac{1}{2},1[f^{\prime }<0$$$$f\left(1\right)=f\left(0\right)=1$$$$\Rightarrow \mathrm{\forall }x\in [0,1]f\left(x\right)⩾1$$$$forx=\frac{1900}{1999}$$$$f\left(\frac{1900}{1999}\right)={\left(\frac{1900}{1999}\right)}^{\frac{3}{4}}+{(1-\frac{1900}{1999})}^{\frac{3}{4}}⩾1$$$$\Rightarrow {\left(1900\right)}^{\frac{3}{4}}+{99}^{\frac{3}{4}}⩾{1999}^{\frac{3}{4}}$$$$$$

Commented by mathdanisur last updated on 05/Aug/21

$$\mathrm{Thank}\mathrm{You}\mathbf{Ser}\mathrm{Cool}$$

Commented by mathdanisur last updated on 06/Aug/21

$$\mathbf{Ser},\mathrm{can}\mathrm{it}\mathrm{be}\mathrm{written}\mathrm{as}{\mathrm{Newton}}^{\prime }\mathrm{s}$$$$\mathrm{Binomial}\mathrm{or}\mathrm{inequality}.?$$

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