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Question Number 205116 by universe last updated on 09/Mar/24

$$\mathrm{let}{\mathrm{x}}^2-3\mathrm{x}+\mathrm{p}=0\mathrm{has}\mathrm{two}\mathrm{positive}\mathrm{roots}$$$${}^{\prime }{\mathrm{a}}^{\prime }\mathrm{and}{}^{\prime }{\mathrm{b}}^{\prime }\mathrm{then}\inf (\frac{4}{\mathrm{a}}+\frac{1}{\mathrm{b}})\mathrm{is}$$

Answered by pi314 last updated on 09/Mar/24

$$\frac{4b+a}{ab}=\frac{3+3b}{p}$$$$9-4p⩾0;p⩽\frac{9}{4}$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$(\frac{4}{a}+\frac{1}{b})=\frac{4}{3-b}+\frac{1}{b}=f\left(b\right)$$$$f^{\prime }\left(b\right)=\frac{4}{{(3-b)}^2}-\frac{1}{b^2}=\frac{4b^2-{(3-b)}^2}{{(3-b)}^2{b}^2}=\frac{3b^2+6b-9}{b^2{(3-b)}^2}$$$$=\frac{3(b^2+2b-3)}{b^2{(3-b)}^2}=\frac{3(b-1)(b+3)}{b^2{(3-b)}^2}$$$$inf\left(f\right)=f\left(1\right)=\frac{4}{2}+1=3$$

Commented by pi314 last updated on 09/Mar/24

$$mistackforme$$

Answered by mr W last updated on 09/Mar/24

$$a,b>0$$$$a+b=3$$$$\frac{4}{a}+\frac{1}{b}=\frac{2^2}{a}+\frac{1^2}{b}⩾\frac{{(2+1)}^2}{a+b}=\frac{9}{3}=3$$

Commented by universe last updated on 09/Mar/24

$$thankyousir$$

Commented by universe last updated on 09/Mar/24

$$sirnameofthisenequality\frac{2^2}{a}+\frac{1^2}{b}⩾\frac{{(2+1)}^2}{a+b}$$

Commented by mr W last updated on 09/Mar/24

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