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Question-195740

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Question Number 195740 by universe last updated on 09/Aug/23
Answered by Frix last updated on 09/Aug/23
Triangle ⇒ a+b>c∧a+c>b∧b+c>a Let b=(u−v)a∧c=(u+v)a ⇒ u>(1/2)∧−(1/2)<v<(1/2) f(u, v)=(1/(2u))+((u−v)/(u+v+1))+((u+v)/(u−v+1)) (3/2)≤f(u, v)<2 (df/dv)=0 ((4(u+1)(2u+1)v)/((u+v+1)^2 (u−v+1)^2 ))=0 ⇒ v=0 f(u, 0)=((4u^2 +u+1)/(2u(u+1))) (df/du)=0 (((u−1)(3u+1))/(2u^2 (u+1)^2 ))=0 ⇒ u=1 f(1, 0)=(3/2) occurs when a=b=c f(u, u)=((4u^2 +1)/(2u)) (df/du)=0 (((2u−1)(2u+1))/(2u^2 ))=0 ⇒ u=(1/2) f((1/2), (1/2))=2 [Limit at a=c∧b=0]
$$\mathrm{Triangle}\:\Rightarrow\:{a}+{b}>{c}\wedge{a}+{c}>{b}\wedge{b}+{c}>{a} \\ $$$$\mathrm{Let}\:{b}=\left({u}−{v}\right){a}\wedge{c}=\left({u}+{v}\right){a} \\ $$$$\Rightarrow\:{u}>\frac{\mathrm{1}}{\mathrm{2}}\wedge−\frac{\mathrm{1}}{\mathrm{2}}<{v}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${f}\left({u},\:{v}\right)=\frac{\mathrm{1}}{\mathrm{2}{u}}+\frac{{u}−{v}}{{u}+{v}+\mathrm{1}}+\frac{{u}+{v}}{{u}−{v}+\mathrm{1}} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\leqslant{f}\left({u},\:{v}\right)<\mathrm{2} \\ $$$$\frac{{df}}{{dv}}=\mathrm{0} \\ $$$$\frac{\mathrm{4}\left({u}+\mathrm{1}\right)\left(\mathrm{2}{u}+\mathrm{1}\right){v}}{\left({u}+{v}+\mathrm{1}\right)^{\mathrm{2}} \left({u}−{v}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0}\:\Rightarrow\:{v}=\mathrm{0} \\ $$$${f}\left({u},\:\mathrm{0}\right)=\frac{\mathrm{4}{u}^{\mathrm{2}} +{u}+\mathrm{1}}{\mathrm{2}{u}\left({u}+\mathrm{1}\right)} \\ $$$$\frac{{df}}{{du}}=\mathrm{0} \\ $$$$\frac{\left({u}−\mathrm{1}\right)\left(\mathrm{3}{u}+\mathrm{1}\right)}{\mathrm{2}{u}^{\mathrm{2}} \left({u}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0}\:\Rightarrow\:{u}=\mathrm{1} \\ $$$${f}\left(\mathrm{1},\:\mathrm{0}\right)=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{occurs}\:\mathrm{when}\:{a}={b}={c} \\ $$$$ \\ $$$${f}\left({u},\:{u}\right)=\frac{\mathrm{4}{u}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{u}} \\ $$$$\frac{{df}}{{du}}=\mathrm{0} \\ $$$$\frac{\left(\mathrm{2}{u}−\mathrm{1}\right)\left(\mathrm{2}{u}+\mathrm{1}\right)}{\mathrm{2}{u}^{\mathrm{2}} }=\mathrm{0}\:\Rightarrow\:{u}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2} \\ $$$$\left[\mathrm{Limit}\:\mathrm{at}\:{a}={c}\wedge{b}=\mathrm{0}\right] \\ $$
Answered by witcher3 last updated on 09/Aug/23
a≥b≥c a+b≥c+a≥b+c Δ=(a/(b+c))+(b/(c+a))+(c/(a+b))≥(b/(b+c))+(c/(c+a))+(a/(a+b)) (a/(b+c))+(b/(c+a))+(c/(a+b))≥(c/(b+c))+(a/(c+a))+(b/(a+b)) 2((a/(b+c))+(b/(c+a))+(c/(a+b)))≥3⇒(a/(b+c))+(b/(c+a))+(c/(a+b))≥(3/2) (a/(b+c))+(b/(a+c))+(c/(a+b))=((2a)/(2b+2c))+((2b)/(2a+2c))+((2c)/(2a+2b)) b+c≥a,b+a≥c,a+c≥b⇒ Δ< ((2a)/(b+a+c))+((2b)/(a+b+c))+((2c)/(a+b+c))=2 (3/2)≤Δ<2
$$\mathrm{a}\geqslant\mathrm{b}\geqslant\mathrm{c} \\ $$$$\mathrm{a}+\mathrm{b}\geqslant\mathrm{c}+\mathrm{a}\geqslant\mathrm{b}+\mathrm{c} \\ $$$$\Delta=\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}\geqslant\frac{\mathrm{b}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{a}}{\mathrm{a}+\mathrm{b}} \\ $$$$\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}\geqslant\frac{\mathrm{c}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{a}}{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{b}}{\mathrm{a}+\mathrm{b}} \\ $$$$\mathrm{2}\left(\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}\right)\geqslant\mathrm{3}\Rightarrow\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}\geqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{a}+\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}=\frac{\mathrm{2a}}{\mathrm{2b}+\mathrm{2c}}+\frac{\mathrm{2b}}{\mathrm{2a}+\mathrm{2c}}+\frac{\mathrm{2c}}{\mathrm{2a}+\mathrm{2b}} \\ $$$$\mathrm{b}+\mathrm{c}\geqslant\mathrm{a},\mathrm{b}+\mathrm{a}\geqslant\mathrm{c},\mathrm{a}+\mathrm{c}\geqslant\mathrm{b}\Rightarrow \\ $$$$\Delta< \\ $$$$\frac{\mathrm{2a}}{\mathrm{b}+\mathrm{a}+\mathrm{c}}+\frac{\mathrm{2b}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}+\frac{\mathrm{2c}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}=\mathrm{2} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\leqslant\Delta<\mathrm{2} \\ $$

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