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Question Number 195740 by universe last updated on 09/Aug/23

Answered by Frix last updated on 09/Aug/23

$$\mathrm{Triangle}\Rightarrow a+b>c\wedge a+c>b\wedge b+c>a$$$$\mathrm{Let}b=(u-v)a\wedge c=(u+v)a$$$$\Rightarrow u>\frac{1}{2}\wedge -\frac{1}{2}<v<\frac{1}{2}$$$$f(u,v)=\frac{1}{2u}+\frac{u-v}{u+v+1}+\frac{u+v}{u-v+1}$$$$\frac{3}{2}⩽f(u,v)<2$$$$\frac{df}{dv}=0$$$$\frac{4(u+1)(2u+1)v}{{(u+v+1)}^2{(u-v+1)}^2}=0\Rightarrow v=0$$$$f(u,0)=\frac{4u^2+u+1}{2u(u+1)}$$$$\frac{df}{du}=0$$$$\frac{(u-1)(3u+1)}{2u^2{(u+1)}^2}=0\Rightarrow u=1$$$$f(1,0)=\frac{3}{2}$$$$\mathrm{occurs}\mathrm{when}a=b=c$$$$$$$$f(u,u)=\frac{4u^2+1}{2u}$$$$\frac{df}{du}=0$$$$\frac{(2u-1)(2u+1)}{2u^2}=0\Rightarrow u=\frac{1}{2}$$$$f(\frac{1}{2},\frac{1}{2})=2$$$$[\mathrm{Limit}\mathrm{at}a=c\wedge b=0]$$

Answered by witcher3 last updated on 09/Aug/23

$$\mathrm{a}⩾\mathrm{b}⩾\mathrm{c}$$$$\mathrm{a}+\mathrm{b}⩾\mathrm{c}+\mathrm{a}⩾\mathrm{b}+\mathrm{c}$$$$\mathrm{\Delta }=\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}⩾\frac{\mathrm{b}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{a}}{\mathrm{a}+\mathrm{b}}$$$$\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}⩾\frac{\mathrm{c}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{a}}{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{b}}{\mathrm{a}+\mathrm{b}}$$$$2(\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}})⩾3\Rightarrow \frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}⩾\frac{3}{2}$$$$\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{a}+\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}=\frac{2\mathrm{a}}{2\mathrm{b}+2\mathrm{c}}+\frac{2\mathrm{b}}{2\mathrm{a}+2\mathrm{c}}+\frac{2\mathrm{c}}{2\mathrm{a}+2\mathrm{b}}$$$$\mathrm{b}+\mathrm{c}⩾\mathrm{a},\mathrm{b}+\mathrm{a}⩾\mathrm{c},\mathrm{a}+\mathrm{c}⩾\mathrm{b}\Rightarrow$$$$\mathrm{\Delta }<$$$$\frac{2\mathrm{a}}{\mathrm{b}+\mathrm{a}+\mathrm{c}}+\frac{2\mathrm{b}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}+\frac{2\mathrm{c}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}=2$$$$\frac{3}{2}⩽\mathrm{\Delta }<2$$

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