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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)
$$\mathrm{1}\leqslant{a};{b};{c};{d}\leqslant\mathrm{2} \\ $$$$\mid\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{b}\right)\left(\mathrm{1}−{c}\right)\left(\mathrm{1}−{d}\right)\mid\leqslant\frac{{abcd}}{\mathrm{4}} \\ $$
Commented by Mathgreat last updated on 17/Dec/20
prove
$$\boldsymbol{{prove}} \\ $$
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)
$$\: \\ $$$$\:\:\:\rightarrow\:\mathrm{1}\leqslant\:{a};{b};{c};{d}\:\leqslant\:\mathrm{2}\:\:\supset \\ $$$$\:\:\:\:\mathrm{1}..\:\:\mathrm{0}\:\leqslant\:{a}−\mathrm{1};\:{b}−\mathrm{1};\:{c}−\mathrm{1};\:{d}−\mathrm{1}\:\leqslant\:\mathrm{1} \\ $$$$\:\:\:\:\mathrm{2}..\:\:−\mathrm{1}\:\leqslant\:\mathrm{1}−{a};\:\mathrm{1}−{b};\:\mathrm{1}−{c};\:\mathrm{1}−{d}\:\leqslant\:\mathrm{0}\:\: \\ $$$$\:\mathrm{numerical}\:\mathrm{values}\:\mathrm{remain}\:\mathrm{the}\:\mathrm{same}\:\left(\mathrm{only}\:\mathrm{sign}\:\mathrm{has}\:\mathrm{changed}\right) \\ $$$$\:\:\:\:\Rightarrow\:\mid\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{b}\right)\left(\mathrm{1}−{c}\right)\left(\mathrm{1}−{d}\right)\mid\leqslant\frac{{abcd}}{\mathrm{4}} \\ $$$$\:\:\:\:\equiv\:\left({a}−\mathrm{1}\right)\left({b}−\mathrm{1}\right)\left({c}−\mathrm{1}\right)\left({d}−\mathrm{1}\right)\:\leqslant\:\frac{{abcd}}{\mathrm{4}} \\ $$$$\:\:\:\: \\ $$$$\:\mathrm{from}\:\mathrm{1}..\:\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:\mathrm{concluded}\:\mathrm{that}\:\mathrm{the}\:\mathrm{maximum} \\ $$$$\:\mathrm{of}\:\left({a}−\mathrm{1}\right)\left({b}−\mathrm{1}\right)\left({c}−\mathrm{1}\right)\left({d}−\mathrm{1}\right)\:\mathrm{is}\:\mathrm{1}\:\mathrm{and}\:\mathrm{that} \\ $$$$\:\mathrm{2}\left({a}−\mathrm{1}\right)\:\leqslant\:{a};\:\:\mathrm{2}\left({b}−\mathrm{1}\right)\:\leqslant\:{b};\:\:\mathrm{2}\left({c}−\mathrm{1}\right)\:\leqslant\:{c};\:\:\mathrm{2}\left({d}−\mathrm{1}\right)\:\leqslant\:{d} \\ $$$$\: \\ $$$$\:\:\:\mathrm{it}\:\mathrm{follows}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2}\left({a}−\mathrm{1}\right)\mathrm{2}\left({b}−\mathrm{1}\right)\mathrm{2}\left({c}−\mathrm{1}\right)\mathrm{2}\left({d}−\mathrm{1}\right)\:\leqslant\:{abcd} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{16}\left({a}−\mathrm{1}\right)\left({b}−\mathrm{1}\right)\left({c}−\mathrm{1}\right)\left({d}−\mathrm{1}\right)\:\leqslant\:{abcd} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{4}\left({a}−\mathrm{1}\right)\left({b}−\mathrm{1}\right)\left({c}−\mathrm{1}\right)\left({d}−\mathrm{1}\right)\:\leqslant\:\frac{{abcd}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\left({a}−\mathrm{1}\right)\left({b}−\mathrm{1}\right)\left({c}−\mathrm{1}\right)\left({d}−\mathrm{1}\right)\:\leqslant\:\mathrm{4}\left({a}−\mathrm{1}\right)\left({b}−\mathrm{1}\right)\left({c}−\mathrm{1}\right)\left({d}−\mathrm{1}\right) \\ $$$$\: \\ $$$$\:\:\Rightarrow\:\left({a}−\mathrm{1}\right)\left({b}−\mathrm{1}\right)\left({c}−\mathrm{1}\right)\left({d}−\mathrm{1}\right)\:\leqslant\:\frac{{abcd}}{\mathrm{4}} \\ $$$$\:\:\therefore\:\:\mid\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{b}\right)\left(\mathrm{1}−{c}\right)\left(\mathrm{1}−{d}\right)\mid\:\leqslant\:\frac{{abcd}}{\mathrm{4}} \\ $$$$\: \\ $$$$\: \\ $$$$\: \\ $$$$\: \\ $$$$\:\:\:\:\: \\ $$

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