Question-98137

Question Number 98137 by Algoritm last updated on 11/Jun/20

Answered by Farruxjano last updated on 11/Jun/20

(1−a)(1−b)(1−c)=(1−(a+b)+ab)(1−c)= =1−c−a−b+ab+ac+ab−abc= =ab+bc+ac−abc=(∗) 1=a+b+c≥3((abc))^(1/3) ⇒ ((abc))^(1/3) ≤(1/3) ab+bc+ac≥3((a^2 b^2 c^2 ))^(1/3) × { ((((abc))^(1/3) ≤(1/3))),((((a^2 b^2 c^2 ))^(1/3) ≤(1/3)(ab+bc+ac))) :}⇒ ⇒abc≤(1/9)(ab+bc+ac) ⇒ ab+bc+ac≥9abc (∗)=ab+bc+ac−abc≥9abc−abc=8abc (1−a)(1−b)(1−c)≥8abc max{k}=8 Solved by: HAMRABOYEV FARRUXJON

$$\left(\mathrm{1}−\boldsymbol{{a}}\right)\left(\mathrm{1}−\boldsymbol{{b}}\right)\left(\mathrm{1}−\boldsymbol{{c}}\right)=\left(\mathrm{1}−\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)+\boldsymbol{{ab}}\right)\left(\mathrm{1}−\boldsymbol{{c}}\right)= \\ $$$$=\mathrm{1}−\boldsymbol{{c}}−\boldsymbol{{a}}−\boldsymbol{{b}}+\boldsymbol{{ab}}+\boldsymbol{{ac}}+\boldsymbol{{ab}}−\boldsymbol{{abc}}= \\ $$$$=\boldsymbol{{ab}}+\boldsymbol{{bc}}+\boldsymbol{{ac}}−\boldsymbol{{abc}}=\left(\ast\right) \\ $$$$\mathrm{1}=\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\boldsymbol{{abc}}}\:\Rightarrow\:\sqrt[{\mathrm{3}}]{\boldsymbol{{abc}}}\leqslant\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\boldsymbol{{ab}}+\boldsymbol{{bc}}+\boldsymbol{{ac}}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{c}}^{\mathrm{2}} } \\ $$$$×\begin{cases}{\sqrt[{\mathrm{3}}]{\boldsymbol{{abc}}}\leqslant\frac{\mathrm{1}}{\mathrm{3}}}\\{\sqrt[{\mathrm{3}}]{\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{c}}^{\mathrm{2}} }\leqslant\frac{\mathrm{1}}{\mathrm{3}}\left(\boldsymbol{{ab}}+\boldsymbol{{bc}}+\boldsymbol{{ac}}\right)}\end{cases}\Rightarrow \\ $$$$\Rightarrow\boldsymbol{{abc}}\leqslant\frac{\mathrm{1}}{\mathrm{9}}\left(\boldsymbol{{ab}}+\boldsymbol{{bc}}+\boldsymbol{{ac}}\right)\:\Rightarrow \\ $$$$\boldsymbol{{ab}}+\boldsymbol{{bc}}+\boldsymbol{{ac}}\geqslant\mathrm{9}\boldsymbol{{abc}} \\ $$$$\left(\ast\right)=\boldsymbol{{ab}}+\boldsymbol{{bc}}+\boldsymbol{{ac}}−\boldsymbol{{abc}}\geqslant\mathrm{9}\boldsymbol{{abc}}−\boldsymbol{{abc}}=\mathrm{8}\boldsymbol{{abc}} \\ $$$$\left(\mathrm{1}−\boldsymbol{{a}}\right)\left(\mathrm{1}−\boldsymbol{{b}}\right)\left(\mathrm{1}−\boldsymbol{{c}}\right)\geqslant\mathrm{8}\boldsymbol{{abc}}\:\:\:\:\boldsymbol{{max}}\left\{\boldsymbol{{k}}\right\}=\mathrm{8} \\ $$$$\boldsymbol{{Solved}}\:\boldsymbol{{by}}:\:\boldsymbol{{HAMRABOYEV}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{FARRUXJON}} \\ $$

Commented by Farruxjano last updated on 11/Jun/20

$$\boldsymbol{{Is}}\:\boldsymbol{{it}}\:\boldsymbol{{right}},\:\boldsymbol{{please}}\:\boldsymbol{{check}}… \\ $$

Commented by Algoritm last updated on 11/Jun/20

$$\mathrm{thanks} \\ $$

Answered by MJS last updated on 11/Jun/20

due to symmetry we have an extreme at a=b=c=(1/3) ((2/3))^3 ≥k((1/3))^3 ⇒ 8≥k

$$\mathrm{due}\:\mathrm{to}\:\mathrm{symmetry}\:\mathrm{we}\:\mathrm{have}\:\mathrm{an}\:\mathrm{extreme}\:\mathrm{at} \\ $$$${a}={b}={c}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} \geqslant{k}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} \\ $$$$\Rightarrow\:\mathrm{8}\geqslant{k} \\ $$

Answered by mathmax by abdo last updated on 11/Jun/20

we have 1−a =b+c ,1−b =a+c ,1−c =a+b ⇒ (1−a)(1−b)(1−c)=(b+c)(a+c)(a+b) also b+c ≥2(√(bc)) a+c ≥2(√(ac)) ,a+b ≥2(√(ab)) ⇒(1−a)(1−b)(1−c)≥8abc ⇒max{k} =8 but this result need more proof..

$$\mathrm{we}\:\mathrm{have}\:\:\mathrm{1}−\mathrm{a}\:=\mathrm{b}+\mathrm{c}\:,\mathrm{1}−\mathrm{b}\:=\mathrm{a}+\mathrm{c}\:,\mathrm{1}−\mathrm{c}\:=\mathrm{a}+\mathrm{b}\:\Rightarrow \\ $$$$\left(\mathrm{1}−\mathrm{a}\right)\left(\mathrm{1}−\mathrm{b}\right)\left(\mathrm{1}−\mathrm{c}\right)=\left(\mathrm{b}+\mathrm{c}\right)\left(\mathrm{a}+\mathrm{c}\right)\left(\mathrm{a}+\mathrm{b}\right)\:\:\mathrm{also}\:\mathrm{b}+\mathrm{c}\:\geqslant\mathrm{2}\sqrt{\mathrm{bc}} \\ $$$$\mathrm{a}+\mathrm{c}\:\geqslant\mathrm{2}\sqrt{\mathrm{ac}}\:,\mathrm{a}+\mathrm{b}\:\geqslant\mathrm{2}\sqrt{\mathrm{ab}}\:\Rightarrow\left(\mathrm{1}−\mathrm{a}\right)\left(\mathrm{1}−\mathrm{b}\right)\left(\mathrm{1}−\mathrm{c}\right)\geqslant\mathrm{8abc}\: \\ $$$$\Rightarrow\mathrm{max}\left\{\mathrm{k}\right\}\:=\mathrm{8}\:\mathrm{but}\:\mathrm{this}\:\mathrm{result}\:\mathrm{need}\:\mathrm{more}\:\mathrm{proof}.. \\ $$

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