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Question Number 210229 by mnjuly1970 last updated on 03/Aug/24
Answered by a.lgnaoui last updated on 03/Aug/24
8=a^3 +b^3 +6ab 16=(a^3 +4)+(b^3 +4)+6ab { ((b^3 +4=16−(a^3 +b^3 )−6ab (1))),(((1/(a^3 +4))+(1/(16−(a^3 +b^3 )−6ab))=x(2))) :} (a^3 +4)+(b^3 +4)=16−6ab a^3 +b^3 =(a+b)(a^2 +b^2 −ab) =2(a^2 +b^2 −ab) ⇒ { ((8−6ab=a^3 +b^3 )),((a^3 +b^3 =2(a^2 +b^2 )−2ab)) :} (1/(a^3 +4))+(1/8)=x=((12+a^3 )/(8(a^3 +4))) (3) a+b=2 b=2−a alors (1/(a^3 +4))+(1/(12−a^3 −2a(2−a)))=x soit −(1/8)+(1/(12−a^3 −2a(2−a)))=0 8−12+a^3 +2a(2−a)=0 a^3 −4+2a(2−a)=0 a^3 −2a^2 +4a−4=0 x=((a^3 −2a^2 +4a−4)/(8−(a^3 −2a^2 +4a−4))) soit 𝛠=a^3 −2a^2 +4a−4 alors x= (𝛠/(8−𝛠)) a+b=2 ⇒ a≤2 { ((a=2 x=((1/2)/(8−1/2))=(1/(15)) b=0)),((a<2 ⇒ )) :} a=1,295597752.. ϱ=0 a=2⇒b=0 (1/(12))+(1/4)=(1/3)=(5/(15)) <((10)/(15)) b=2 a=0 (1/4)+(1/8)=(3/8)=((15)/(40))≤((16)/(40))=(2/5) or (1/3) <(3/8) { (((1/3)= minimum relatif)),(((3/8)=maximum relatif)) :} ⇒ (1/3)<(1/(a^3 +4))+(1/(b^3 +4))<(3/8) ≤(2/5) autre methode (3)⇒ (dx/da)=(1/8)(((3a^2 (4+a^3 )−(12+a^3 )3a^2 )/((4+a^3 )^2 ))) ⇒3a^2 /(4+a^3 )^2 max=((24a^2 )/((4+a^3 )^2 )) ⇒ a=0 soit b=2 alors (1/(a^3 +4))+(1/(b^3 +4))=(3/8)=((15)/(40))≤((16)/(40))=(2/5)
$$\mathrm{8}=\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\boldsymbol{\mathrm{b}}^{\mathrm{3}} +\mathrm{6}\boldsymbol{\mathrm{ab}} \\ $$$$ \\ $$$$\mathrm{16}=\left(\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\mathrm{4}\right)+\left(\boldsymbol{\mathrm{b}}^{\mathrm{3}} +\mathrm{4}\right)+\mathrm{6}\boldsymbol{\mathrm{ab}}\:\:\: \\ $$$$\begin{cases}{\mathrm{b}^{\mathrm{3}} +\mathrm{4}=\mathrm{16}−\left(\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} \right)−\mathrm{6ab}\:\:\:\:\:\:\left(\mathrm{1}\right)}\\{\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} +\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{16}−\left(\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} \right)−\mathrm{6ab}}=\mathrm{x}\left(\mathrm{2}\right)}\end{cases} \\ $$$$ \\ $$$$\left(\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\mathrm{4}\right)+\left(\boldsymbol{\mathrm{b}}^{\mathrm{3}} +\mathrm{4}\right)=\mathrm{16}−\mathrm{6}\boldsymbol{\mathrm{ab}} \\ $$$$\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\boldsymbol{\mathrm{b}}^{\mathrm{3}} =\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right)\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} −\boldsymbol{\mathrm{ab}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} −\boldsymbol{\mathrm{ab}}\right) \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{8}−\mathrm{6ab}=\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} }\\{\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} =\mathrm{2}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)−\mathrm{2ab}}\end{cases} \\ $$$$\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} +\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}=\mathrm{x}=\frac{\mathrm{12}+\mathrm{a}^{\mathrm{3}} }{\mathrm{8}\left(\mathrm{a}^{\mathrm{3}} +\mathrm{4}\right)}\:\:\:\:\:\left(\mathrm{3}\right) \\ $$$$\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}=\mathrm{2}\:\:\:\:\:\:\:\boldsymbol{\mathrm{b}}=\mathrm{2}−\boldsymbol{\mathrm{a}} \\ $$$$\:\:\boldsymbol{\mathrm{alors}}\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{12}−\boldsymbol{\mathrm{a}}^{\mathrm{3}} −\mathrm{2}\boldsymbol{\mathrm{a}}\left(\mathrm{2}−\boldsymbol{\mathrm{a}}\right)}=\boldsymbol{\mathrm{x}} \\ $$$$ \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{soit}}\:\: \\ $$$$\:\:−\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{12}−\boldsymbol{\mathrm{a}}^{\mathrm{3}} −\mathrm{2}\boldsymbol{\mathrm{a}}\left(\mathrm{2}−\boldsymbol{\mathrm{a}}\right)}=\mathrm{0} \\ $$$$\mathrm{8}−\mathrm{12}+\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\mathrm{2}\boldsymbol{\mathrm{a}}\left(\mathrm{2}−\boldsymbol{\mathrm{a}}\right)=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{a}}^{\mathrm{3}} −\mathrm{4}+\mathrm{2}\boldsymbol{\mathrm{a}}\left(\mathrm{2}−\boldsymbol{\mathrm{a}}\right)=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{a}}^{\mathrm{3}} −\mathrm{2}\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{\mathrm{a}}−\mathrm{4}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{x}}=\frac{\boldsymbol{\mathrm{a}}^{\mathrm{3}} −\mathrm{2}\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{\mathrm{a}}−\mathrm{4}}{\mathrm{8}−\left(\boldsymbol{\mathrm{a}}^{\mathrm{3}} −\mathrm{2}\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{\mathrm{a}}−\mathrm{4}\right)} \\ $$$$\boldsymbol{\mathrm{soit}}\:\boldsymbol{\varrho}=\boldsymbol{\mathrm{a}}^{\mathrm{3}} −\mathrm{2}\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{\mathrm{a}}−\mathrm{4} \\ $$$$\boldsymbol{\mathrm{alors}}\:\:\boldsymbol{\mathrm{x}}=\:\frac{\boldsymbol{\varrho}}{\mathrm{8}−\boldsymbol{\varrho}}\: \\ $$$$\mathrm{a}+\mathrm{b}=\mathrm{2}\:\:\:\:\Rightarrow\:\:\:\mathrm{a}\leqslant\mathrm{2} \\ $$$$\begin{cases}{\mathrm{a}=\mathrm{2}\:\:\:\:\:\:\mathrm{x}=\frac{\mathrm{1}/\mathrm{2}}{\mathrm{8}−\mathrm{1}/\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{15}}\:\:\:\mathrm{b}=\mathrm{0}}\\{\mathrm{a}<\mathrm{2}\:\:\Rightarrow\:\:}\end{cases} \\ $$$$\:\mathrm{a}=\mathrm{1},\mathrm{295597752}..\:\:\:\:\varrho=\mathrm{0} \\ $$$$\mathrm{a}=\mathrm{2}\Rightarrow\mathrm{b}=\mathrm{0}\: \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{5}}{\mathrm{15}}\:\:\:\:<\frac{\mathrm{10}}{\mathrm{15}} \\ $$$$\mathrm{b}=\mathrm{2}\:\:\:\mathrm{a}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}=\frac{\mathrm{3}}{\mathrm{8}}=\frac{\mathrm{15}}{\mathrm{40}}\leqslant\frac{\mathrm{16}}{\mathrm{40}}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\mathrm{or}\:\:\frac{\mathrm{1}}{\mathrm{3}}\:<\frac{\mathrm{3}}{\mathrm{8}}\:\:\:\:\:\:\begin{cases}{\frac{\mathrm{1}}{\mathrm{3}}=\:\:\mathrm{minimum}\:\mathrm{relatif}}\\{\frac{\mathrm{3}}{\mathrm{8}}=\mathrm{maximum}\:\mathrm{relatif}}\end{cases}\: \\ $$$$\:\:\Rightarrow\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{3}}<\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} +\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{3}} +\mathrm{4}}<\frac{\mathrm{3}}{\mathrm{8}}\:\:\leqslant\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\mathrm{autre}\:\mathrm{methode} \\ $$$$\left(\mathrm{3}\right)\Rightarrow\:\frac{\mathrm{dx}}{\mathrm{da}}=\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\mathrm{3a}^{\mathrm{2}} \left(\mathrm{4}+\mathrm{a}^{\mathrm{3}} \right)−\left(\mathrm{12}+\mathrm{a}^{\mathrm{3}} \right)\mathrm{3a}^{\mathrm{2}} }{\left(\mathrm{4}+\mathrm{a}^{\mathrm{3}} \right)^{\mathrm{2}} }\right) \\ $$$$\:\:\: \\ $$$$\Rightarrow\mathrm{3a}^{\mathrm{2}} /\left(\mathrm{4}+\mathrm{a}^{\mathrm{3}} \right)^{\mathrm{2}} \mathrm{max}=\frac{\mathrm{24a}^{\mathrm{2}} }{\left(\mathrm{4}+\mathrm{a}^{\mathrm{3}} \right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\boldsymbol{\mathrm{a}}=\mathrm{0}\:\boldsymbol{\mathrm{soit}}\:\:\boldsymbol{\mathrm{b}}=\mathrm{2} \\ $$$$\boldsymbol{\mathrm{alors}} \\ $$$$\:\: \\ $$$$\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\mathrm{4}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{b}}^{\mathrm{3}} +\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{8}}=\frac{\mathrm{15}}{\mathrm{40}}\leqslant\frac{\mathrm{16}}{\mathrm{40}}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\:\:\: \\ $$
Answered by Frix last updated on 04/Aug/24
b=2−a; 0≤a≤2 (1/(a^3 +4))+(1/(b^3 +4))= =−((2(3a^2 −6a+8))/(a^6 +6a^5 +12a^4 −8a^3 −24a^2 +48a−48))= =−((2(3a(2−a)−8))/(a^3 (2−a)^3 −24a(2−a)+48)) 0≤a≤2 ⇒ 0≤a(2−a)≤1 t=a(2−a), 0≤t≤1 ⇒ (1/3)≤−((2(3t−8))/(t^3 −24t+48))≤(2/5)
$${b}=\mathrm{2}−{a};\:\mathrm{0}\leqslant{a}\leqslant\mathrm{2} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{3}} +\mathrm{4}}+\frac{\mathrm{1}}{{b}^{\mathrm{3}} +\mathrm{4}}= \\ $$$$=−\frac{\mathrm{2}\left(\mathrm{3}{a}^{\mathrm{2}} −\mathrm{6}{a}+\mathrm{8}\right)}{{a}^{\mathrm{6}} +\mathrm{6}{a}^{\mathrm{5}} +\mathrm{12}{a}^{\mathrm{4}} −\mathrm{8}{a}^{\mathrm{3}} −\mathrm{24}{a}^{\mathrm{2}} +\mathrm{48}{a}−\mathrm{48}}= \\ $$$$=−\frac{\mathrm{2}\left(\mathrm{3}{a}\left(\mathrm{2}−{a}\right)−\mathrm{8}\right)}{{a}^{\mathrm{3}} \left(\mathrm{2}−{a}\right)^{\mathrm{3}} −\mathrm{24}{a}\left(\mathrm{2}−{a}\right)+\mathrm{48}} \\ $$$$\mathrm{0}\leqslant{a}\leqslant\mathrm{2}\:\Rightarrow\:\mathrm{0}\leqslant{a}\left(\mathrm{2}−{a}\right)\leqslant\mathrm{1} \\ $$$${t}={a}\left(\mathrm{2}−{a}\right),\:\mathrm{0}\leqslant{t}\leqslant\mathrm{1}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\leqslant−\frac{\mathrm{2}\left(\mathrm{3}{t}−\mathrm{8}\right)}{{t}^{\mathrm{3}} −\mathrm{24}{t}+\mathrm{48}}\leqslant\frac{\mathrm{2}}{\mathrm{5}} \\ $$
Answered by A5T last updated on 05/Aug/24
(d/da)((1/(a^3 +4))+(1/((2−a)^3 +4)))=((−3a^2 )/((a^3 +4)^2 ))+((3(2−a)^2 )/([(2−a)^3 +4]^2 ))=0 ⇒(((a(√3))^2 )/([(√3)(2−a)]^2 ))=(((a^3 +4)^2 )/([(2−a)^3 +4]^2 ))⇒(a/(2−a))=+_− ((a^3 +4)/((2−a)^3 +4)) Case I: a(2−a)^3 +4a=(2−a)a^3 +4(2−a) ⇒a(2−a)[(2−a)^2 −a^2 ]=−4a+8−4a ⇒a(2−a)(4−4a)=8(1−a)⇒a=1 or a^2 −2a+2=0 ⇒Equality and maximum holds at a=1⇒b=1
$$\frac{{d}}{{da}}\left(\frac{\mathrm{1}}{{a}^{\mathrm{3}} +\mathrm{4}}+\frac{\mathrm{1}}{\left(\mathrm{2}−{a}\right)^{\mathrm{3}} +\mathrm{4}}\right)=\frac{−\mathrm{3}{a}^{\mathrm{2}} }{\left({a}^{\mathrm{3}} +\mathrm{4}\right)^{\mathrm{2}} }+\frac{\mathrm{3}\left(\mathrm{2}−{a}\right)^{\mathrm{2}} }{\left[\left(\mathrm{2}−{a}\right)^{\mathrm{3}} +\mathrm{4}\right]^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\frac{\left({a}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }{\left[\sqrt{\mathrm{3}}\left(\mathrm{2}−{a}\right)\right]^{\mathrm{2}} }=\frac{\left({a}^{\mathrm{3}} +\mathrm{4}\right)^{\mathrm{2}} }{\left[\left(\mathrm{2}−{a}\right)^{\mathrm{3}} +\mathrm{4}\right]^{\mathrm{2}} }\Rightarrow\frac{{a}}{\mathrm{2}−{a}}=\underset{−} {+}\frac{{a}^{\mathrm{3}} +\mathrm{4}}{\left(\mathrm{2}−{a}\right)^{\mathrm{3}} +\mathrm{4}} \\ $$$${Case}\:{I}:\:{a}\left(\mathrm{2}−{a}\right)^{\mathrm{3}} +\mathrm{4}{a}=\left(\mathrm{2}−{a}\right){a}^{\mathrm{3}} +\mathrm{4}\left(\mathrm{2}−{a}\right) \\ $$$$\Rightarrow{a}\left(\mathrm{2}−{a}\right)\left[\left(\mathrm{2}−{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} \right]=−\mathrm{4}{a}+\mathrm{8}−\mathrm{4}{a} \\ $$$$\Rightarrow{a}\left(\mathrm{2}−{a}\right)\left(\mathrm{4}−\mathrm{4}{a}\right)=\mathrm{8}\left(\mathrm{1}−{a}\right)\Rightarrow{a}=\mathrm{1}\:{or}\:{a}^{\mathrm{2}} −\mathrm{2}{a}+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow{Equality}\:{and}\:{maximum}\:{holds}\:{at}\:{a}=\mathrm{1}\Rightarrow{b}=\mathrm{1} \\ $$

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