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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

$$8={\mathbf{a}}^3+{\mathbf{b}}^3+6\mathbf{ab}$$$$$$$$16=({\mathbf{a}}^3+4)+({\mathbf{b}}^3+4)+6\mathbf{ab}$$$$\{\begin{array}{l}{\mathrm{b}}^3+4=16-({\mathrm{a}}^3+{\mathrm{b}}^3)-6\mathrm{ab}\left(1\right)\\ \frac{1}{{\mathrm{a}}^3+4}+\frac{1}{16-({\mathrm{a}}^3+{\mathrm{b}}^3)-6\mathrm{ab}}=\mathrm{x}\left(2\right)\end{array}$$$$$$$$({\mathbf{a}}^3+4)+({\mathbf{b}}^3+4)=16-6\mathbf{ab}$$$${\mathbf{a}}^3+{\mathbf{b}}^3=(\mathbf{a}+\mathbf{b})({\mathbf{a}}^2+{\mathbf{b}}^2-\mathbf{ab})$$$$=2({\mathbf{a}}^2+{\mathbf{b}}^2-\mathbf{ab})$$$$\Rightarrow \{\begin{array}{l}8-6\mathrm{ab}={\mathrm{a}}^3+{\mathrm{b}}^3\\ {\mathrm{a}}^3+{\mathrm{b}}^3=2({\mathrm{a}}^2+{\mathrm{b}}^2)-2\mathrm{ab}\end{array}$$$$\frac{1}{{\mathrm{a}}^3+4}+\frac{1}{8}=\mathrm{x}=\frac{12+{\mathrm{a}}^3}{8({\mathrm{a}}^3+4)}\left(3\right)$$$$\mathbf{a}+\mathbf{b}=2\mathbf{b}=2-\mathbf{a}$$$$\mathbf{alors}\frac{1}{{\mathbf{a}}^3+4}+\frac{1}{12-{\mathbf{a}}^3-2\mathbf{a}(2-\mathbf{a})}=\mathbf{x}$$$$$$$$\mathbf{soit}$$$$-\frac{1}{8}+\frac{1}{12-{\mathbf{a}}^3-2\mathbf{a}(2-\mathbf{a})}=0$$$$8-12+{\mathbf{a}}^3+2\mathbf{a}(2-\mathbf{a})=0$$$${\mathbf{a}}^3-4+2\mathbf{a}(2-\mathbf{a})=0$$$${\mathbf{a}}^3-2{\mathbf{a}}^2+4\mathbf{a}-4=0$$$$\mathbf{x}=\frac{{\mathbf{a}}^3-2{\mathbf{a}}^2+4\mathbf{a}-4}{8-({\mathbf{a}}^3-2{\mathbf{a}}^2+4\mathbf{a}-4)}$$$$\mathbf{soit}\mathit{\varrho }={\mathbf{a}}^3-2{\mathbf{a}}^2+4\mathbf{a}-4$$$$\mathbf{alors}\mathbf{x}=\frac{\mathit{\varrho }}{8-\mathit{\varrho }}$$$$\mathrm{a}+\mathrm{b}=2\Rightarrow \mathrm{a}⩽2$$$$\{\begin{array}{l}\mathrm{a}=2\mathrm{x}=\frac{1/2}{8-1/2}=\frac{1}{15}\mathrm{b}=0\\ \mathrm{a}<2\Rightarrow \end{array}$$$$\mathrm{a}=1,295597752..\varrho =0$$$$\mathrm{a}=2\Rightarrow \mathrm{b}=0$$$$\frac{1}{12}+\frac{1}{4}=\frac{1}{3}=\frac{5}{15}<\frac{10}{15}$$$$\mathrm{b}=2\mathrm{a}=0$$$$\frac{1}{4}+\frac{1}{8}=\frac{3}{8}=\frac{15}{40}⩽\frac{16}{40}=\frac{2}{5}$$$$\mathrm{or}\frac{1}{3}<\frac{3}{8}\{\begin{array}{l}\frac{1}{3}=\mathrm{minimum}\mathrm{relatif}\\ \frac{3}{8}=\mathrm{maximum}\mathrm{relatif}\end{array}$$$$\Rightarrow \frac{1}{3}<\frac{1}{{\mathrm{a}}^3+4}+\frac{1}{{\mathrm{b}}^3+4}<\frac{3}{8}⩽\frac{2}{5}$$$$\mathrm{autre}\mathrm{methode}$$$$\left(3\right)\Rightarrow \frac{\mathrm{dx}}{\mathrm{da}}=\frac{1}{8}\left(\frac{{3\mathrm{a}}^2(4+{\mathrm{a}}^3)-(12+{\mathrm{a}}^3){3\mathrm{a}}^2}{{(4+{\mathrm{a}}^3)}^2}\right)$$$$$$$$\Rightarrow {3\mathrm{a}}^2/{(4+{\mathrm{a}}^3)}^2\max =\frac{{24\mathrm{a}}^2}{{(4+{\mathrm{a}}^3)}^2}\Rightarrow$$$$\mathbf{a}=0\mathbf{soit}\mathbf{b}=2$$$$\mathbf{alors}$$$$$$$$\frac{1}{{\mathbf{a}}^3+4}+\frac{1}{{\mathbf{b}}^3+4}=\frac{3}{8}=\frac{15}{40}⩽\frac{16}{40}=\frac{2}{5}$$$$$$

Answered by Frix last updated on 04/Aug/24

$$b=2-a;0⩽a⩽2$$$$\frac{1}{a^3+4}+\frac{1}{b^3+4}=$$$$=-\frac{2(3a^2-6a+8)}{a^6+6a^5+12a^4-8a^3-24a^2+48a-48}=$$$$=-\frac{2(3a(2-a)-8)}{a^3{(2-a)}^3-24a(2-a)+48}$$$$0⩽a⩽2\Rightarrow 0⩽a(2-a)⩽1$$$$t=a(2-a),0⩽t⩽1\Rightarrow$$$$\frac{1}{3}⩽-\frac{2(3t-8)}{t^3-24t+48}⩽\frac{2}{5}$$

Answered by A5T last updated on 05/Aug/24

$$\frac{d}{da}(\frac{1}{a^3+4}+\frac{1}{{(2-a)}^3+4})=\frac{-3a^2}{{(a^3+4)}^2}+\frac{3{(2-a)}^2}{{[{(2-a)}^3+4]}^2}=0$$$$\Rightarrow \frac{{\left(a\sqrt{3}\right)}^2}{{\left[\sqrt{3}(2-a)\right]}^2}=\frac{{(a^3+4)}^2}{{[{(2-a)}^3+4]}^2}\Rightarrow \frac{a}{2-a}=\underset{-}{+}\frac{a^3+4}{{(2-a)}^3+4}$$$$CaseI:a{(2-a)}^3+4a=(2-a)a^3+4(2-a)$$$$\Rightarrow a(2-a)[{(2-a)}^2-a^2]=-4a+8-4a$$$$\Rightarrow a(2-a)(4-4a)=8(1-a)\Rightarrow a=1or{a}^2-2a+2=0$$$$\Rightarrow Equalityandmaximumholdsata=1\Rightarrow b=1$$

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