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Question Number 191301 by Mingma last updated on 22/Apr/23
Answered by a.lgnaoui last updated on 23/Apr/23
a;b;c>0 ⇒(a+b+c>0 et abc>0) poxons x=abc y=a+b+c xy=1 a+b=y−c a+c=y−b (y−c)(y−b) =y^2 −(b+c)y+bc ⇒ay+(x/a) xy=1⇒ x=(1/y) ay+(1/(ay)) ay=z (a+c)(a+b)= z+(1/z) Extremum de P(z) est: ((d(P(z))/dz)=0 1−(1/z^2 )=0 z>0 ⇒z=1 ay=1 y=(1/a) P(z)=2 Donc minimum de( a+b)(a+c) est: 2
$$\mathrm{a};\mathrm{b};\mathrm{c}>\mathrm{0}\:\:\:\Rightarrow\left(\mathrm{a}+\mathrm{b}+\mathrm{c}>\mathrm{0}\:\:\:\mathrm{et}\:\mathrm{abc}>\mathrm{0}\right) \\ $$$$\mathrm{poxons}\:\:\mathrm{x}=\mathrm{abc}\:\:\:\:\:\mathrm{y}=\mathrm{a}+\mathrm{b}+\mathrm{c}\:\:\:\mathrm{xy}=\mathrm{1} \\ $$$$\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{y}}−\boldsymbol{\mathrm{c}}\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{c}}=\boldsymbol{\mathrm{y}}−\boldsymbol{\mathrm{b}} \\ $$$$\:\:\left(\boldsymbol{\mathrm{y}}−\boldsymbol{\mathrm{c}}\right)\left(\boldsymbol{\mathrm{y}}−\boldsymbol{\mathrm{b}}\right)\:=\boldsymbol{\mathrm{y}}^{\mathrm{2}} −\left(\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{bc}} \\ $$$$\Rightarrow\boldsymbol{\mathrm{ay}}+\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{a}}}\:\:\:\:\:\:\:\boldsymbol{\mathrm{xy}}=\mathrm{1}\Rightarrow\:\:\:\boldsymbol{\mathrm{x}}=\frac{\mathrm{1}}{\boldsymbol{\mathrm{y}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{ay}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{ay}}} \\ $$$$\boldsymbol{\mathrm{ay}}=\boldsymbol{\mathrm{z}}\:\:\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{c}}\right)\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right)=\:\:\:\boldsymbol{\mathrm{z}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{z}}} \\ $$$$\boldsymbol{\mathrm{Extremum}}\:\boldsymbol{\mathrm{de}}\:\boldsymbol{\mathrm{P}}\left(\boldsymbol{\mathrm{z}}\right)\:\boldsymbol{\mathrm{est}}: \\ $$$$\frac{\boldsymbol{\mathrm{d}}\left(\boldsymbol{\mathrm{P}}\left(\boldsymbol{\mathrm{z}}\right)\right.}{\boldsymbol{\mathrm{dz}}}=\mathrm{0}\:\:\:\:\mathrm{1}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{z}}^{\mathrm{2}} }=\mathrm{0}\:\:\:\mathrm{z}>\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\boldsymbol{\mathrm{z}}=\mathrm{1} \\ $$$$\boldsymbol{\mathrm{ay}}=\mathrm{1}\:\:\:\:\:\:\boldsymbol{\mathrm{y}}=\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}}\:\:\:\boldsymbol{\mathrm{P}}\left(\boldsymbol{\mathrm{z}}\right)=\mathrm{2} \\ $$$$\boldsymbol{\mathrm{Donc}}\:\:\boldsymbol{\mathrm{minimum}}\:\boldsymbol{\mathrm{de}}\left(\:\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right)\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{c}}\right)\:\:\boldsymbol{\mathrm{est}}:\:\:\mathrm{2} \\ $$
Answered by mr W last updated on 23/Apr/23
(a+b)(a+c) =(a+b+c−c)(a+b+c−b) =((1/(abc))−c)((1/(abc))−b) =((1/(abc)))^2 −(b+c)(1/(abc))+bc =((1/(abc)))^2 −(a+b+c−a)(1/(abc))+bc =((1/(abc)))^2 −((1/(abc))−a)(1/(abc))+bc =(1/(bc))+bc ≥2(√((1/(bc))×bc))=2 ⇒minimum =2
$$\left({a}+{b}\right)\left({a}+{c}\right) \\ $$$$=\left({a}+{b}+{c}−{c}\right)\left({a}+{b}+{c}−{b}\right) \\ $$$$=\left(\frac{\mathrm{1}}{{abc}}−{c}\right)\left(\frac{\mathrm{1}}{{abc}}−{b}\right) \\ $$$$=\left(\frac{\mathrm{1}}{{abc}}\right)^{\mathrm{2}} −\left({b}+{c}\right)\frac{\mathrm{1}}{{abc}}+{bc} \\ $$$$=\left(\frac{\mathrm{1}}{{abc}}\right)^{\mathrm{2}} −\left({a}+{b}+{c}−{a}\right)\frac{\mathrm{1}}{{abc}}+{bc} \\ $$$$=\left(\frac{\mathrm{1}}{{abc}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{{abc}}−{a}\right)\frac{\mathrm{1}}{{abc}}+{bc} \\ $$$$=\frac{\mathrm{1}}{{bc}}+{bc} \\ $$$$\geqslant\mathrm{2}\sqrt{\frac{\mathrm{1}}{{bc}}×{bc}}=\mathrm{2} \\ $$$$\Rightarrow{minimum}\:=\mathrm{2} \\ $$

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