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1-a-b-c-d-2-1-a-1-b-1-c-1-d-abcd-4-

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Question Number 126105 by Mathgreat last updated on 17/Dec/20
1≤a;b;c;d≤2 ∣(1−a)(1−b)(1−c)(1−d)∣≤((abcd)/4)
1⩽a;b;c;d⩽2∣(1−a)(1−b)(1−c)(1−d)∣⩽abcd4
Commented by Mathgreat last updated on 17/Dec/20
prove
prove
Answered by talminator2856791 last updated on 17/Dec/20
→ 1≤ a;b;c;d ≤ 2 ⊃ 1.. 0 ≤ a−1; b−1; c−1; d−1 ≤ 1 2.. −1 ≤ 1−a; 1−b; 1−c; 1−d ≤ 0 numerical values remain the same (only sign has changed) ⇒ ∣(1−a)(1−b)(1−c)(1−d)∣≤((abcd)/4) ≡ (a−1)(b−1)(c−1)(d−1) ≤ ((abcd)/4) from 1.. it can be concluded that the maximum of (a−1)(b−1)(c−1)(d−1) is 1 and that 2(a−1) ≤ a; 2(b−1) ≤ b; 2(c−1) ≤ c; 2(d−1) ≤ d it follows that 2(a−1)2(b−1)2(c−1)2(d−1) ≤ abcd 16(a−1)(b−1)(c−1)(d−1) ≤ abcd 4(a−1)(b−1)(c−1)(d−1) ≤ ((abcd)/4) (a−1)(b−1)(c−1)(d−1) ≤ 4(a−1)(b−1)(c−1)(d−1) ⇒ (a−1)(b−1)(c−1)(d−1) ≤ ((abcd)/4) ∴ ∣(1−a)(1−b)(1−c)(1−d)∣ ≤ ((abcd)/4)
→1⩽a;b;c;d⩽2⊃1..0⩽a−1;b−1;c−1;d−1⩽12..−1⩽1−a;1−b;1−c;1−d⩽0numericalvaluesremainthesame(onlysignhaschanged)⇒∣(1−a)(1−b)(1−c)(1−d)∣⩽abcd4≡(a−1)(b−1)(c−1)(d−1)⩽abcd4from1..itcanbeconcludedthatthemaximumof(a−1)(b−1)(c−1)(d−1)is1andthat2(a−1)⩽a;2(b−1)⩽b;2(c−1)⩽c;2(d−1)⩽ditfollowsthat2(a−1)2(b−1)2(c−1)2(d−1)⩽abcd16(a−1)(b−1)(c−1)(d−1)⩽abcd4(a−1)(b−1)(c−1)(d−1)⩽abcd4(a−1)(b−1)(c−1)(d−1)⩽4(a−1)(b−1)(c−1)(d−1)⇒(a−1)(b−1)(c−1)(d−1)⩽abcd4∴∣(1−a)(1−b)(1−c)(1−d)∣⩽abcd4

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