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

	Answered by Farruxjano last updated on 11/Jun/20
	$(1−a)(1−b)(1−c)=(1−(a+b)+ab)(1−c)= =1−c−a−b+ab+ac+ab−abc= =ab+bc+ac−abc=(∗) 1=a+b+c≥3((abc))^(1/3) ⇒ ((abc))^(1/3) ≤(1/3) ab+bc+ac≥3((a^2 b^2 c^2 ))^(1/3) × { ((((abc))^(1/3) ≤(1/3))),((((a^2 b^2 c^2 ))^(1/3) ≤(1/3)(ab+bc+ac))) :}⇒ ⇒abc≤(1/9)(ab+bc+ac) ⇒ ab+bc+ac≥9abc (∗)=ab+bc+ac−abc≥9abc−abc=8abc (1−a)(1−b)(1−c)≥8abc max{k}=8 Solved by: HAMRABOYEV FARRUXJON$
	$(1 - a) (1 - b) (1 - c) = (1 - (a + b) + a b) (1 - c) =$ $= 1 - c - a - b + a b + a c + a b - a b c =$ $= a b + b c + a c - a b c = ()$ $1 = a + b + c ⩾ 3 \sqrt[3]{a b c} \Rightarrow \sqrt[3]{a b c} ⩽ \frac{1}{3}$ $a b + b c + a c ⩾ 3 \sqrt[3]{a^{2} b^{2} c^{2}}$ $\times {\begin{cases} \sqrt[3]{a b c} ⩽ \frac{1}{3} \\ \sqrt[3]{a^{2} b^{2} c^{2}} ⩽ \frac{1}{3} (a b + b c + a c) \end{cases} \Rightarrow$ $\Rightarrow a b c ⩽ \frac{1}{9} (a b + b c + a c) \Rightarrow$ $a b + b c + a c ⩾ 9 a b c$ $() = a b + b c + a c - a b c ⩾ 9 a b c - a b c = 8 a b c$ $(1 - a) (1 - b) (1 - c) ⩾ 8 a b c m a x {k} = 8$ $S o l v e d b y : H A M R A B O Y E V$ $F A R R U X J O N$
	Commented by Farruxjano last updated on 11/Jun/20

	$I s i t r i g h t, p l e a s e c h e c k . . .$
	Commented by Algoritm last updated on 11/Jun/20

	$thanks$
	Answered by MJS last updated on 11/Jun/20

	$due to symmetry we have an extreme at$ $a = b = c = \frac{1}{3}$ ${(\frac{2}{3})}^{3} ⩾ k {(\frac{1}{3})}^{3}$ $\Rightarrow 8 ⩾ k$
	Answered by mathmax by abdo last updated on 11/Jun/20

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