Track / Stage 8 / 58 of 180 #1758 of 2000
Problem 1758 IMO Shortlist mid-range; USAMO P2/P5 Difficulty 8.2 Prove it 2020 Taiwan IMO 1J · Taiwan · 2020
設實數 a , b , c , d a, b, c, d a , b , c , d ( a + c ) ( b + d ) = 2 ( a c − 2 b d − 1 ) .
(a + c)(b + d) = \sqrt{2}(ac - 2bd - 1).
( a + c ) ( b + d ) = 2 ( a c − 2 b d − 1 ) . ( a b − 1 ) 2 + ( b c − 1 ) 2 + ( c d − 1 ) 2 + ( d a − 1 ) 2 + ( a c − 1 ) 2 + ( 2 b d + 1 ) 2 ≥ 4.
(ab - 1)^2 + (bc - 1)^2 + (cd - 1)^2 + (da - 1)^2 + (ac - 1)^2 + (2bd + 1)^2 \geq 4.
( ab − 1 ) 2 + ( b c − 1 ) 2 + ( c d − 1 ) 2 + ( d a − 1 ) 2 + ( a c − 1 ) 2 + ( 2 b d + 1 ) 2 ≥ 4.
Let a , b , c , d a, b, c, d a , b , c , d ( a + c ) ( b + d ) = 2 ( a c − 2 b d − 1 ) .
(a + c)(b + d) = \sqrt{2}(ac - 2bd - 1).
( a + c ) ( b + d ) = 2 ( a c − 2 b d − 1 ) . ( a b − 1 ) 2 + ( b c − 1 ) 2 + ( c d − 1 ) 2 + ( d a − 1 ) 2 + ( a c − 1 ) 2 + ( 2 b d + 1 ) 2 ≥ 4.
(ab - 1)^2 + (bc - 1)^2 + (cd - 1)^2 + (da - 1)^2 + (ac - 1)^2 + (2bd + 1)^2 \geq 4.
( ab − 1 ) 2 + ( b c − 1 ) 2 + ( c d − 1 ) 2 + ( d a − 1 ) 2 + ( a c − 1 ) 2 + ( 2 b d + 1 ) 2 ≥ 4.
This one wants a proof. Work it on paper, read the official solution, then mark
yourself honestly — the ladder only means something if the record is true.
I solved it I didn't Skip
Official solution 令 A = ( a + c ) ( b + d ) = 2 ( a c − 2 b d − 1 ) A = (a+c)(b+d) = \sqrt{2}(ac-2bd-1) A = ( a + c ) ( b + d ) = 2 ( a c − 2 b d − 1 ) ∑ cyc ( a b − 1 ) 2 ≥ ∑ cyc ( a b − 1 ) 2 − ( ∑ cyc a b − 1 ) 2 = − 2 ( ∑ cyc a b 2 c ) − 4 a b c d + 3 = 3 − 4 a b c d − 2 ( a b + c d ) ( a d + b c ) = 3 − 4 a b c d − 2 ( a b + c d ) ( A − a b − c d ) = 3 − 4 a b c d + 2 ( a b + c d − A 2 ) 2 − A 2 2 ≥ 3 − 4 a b c d − ( a c − 2 b d − 1 ) 2 = 4 − ( a c − 1 ) 2 − ( 2 b d + 1 ) 2
\begin{align*}
\sum_{\text{cyc}} (ab-1)^2 &\ge \sum_{\text{cyc}} (ab-1)^2 - \left(\sum_{\text{cyc}} ab-1\right)^2 \\
&= -2 \left(\sum_{\text{cyc}} ab^2 c\right) - 4abcd + 3 \\
&= 3 - 4abcd - 2(ab + cd)(ad + bc) \\
&= 3 - 4abcd - 2(ab + cd)(A - ab - cd) \\
&= 3 - 4abcd + 2 \left(ab + cd - \frac{A}{2}\right)^2 - \frac{A^2}{2} \\
&\ge 3 - 4abcd - (ac - 2bd - 1)^2 \\
&= 4 - (ac - 1)^2 - (2bd + 1)^2
\end{align*}
cyc ∑ ( ab − 1 ) 2 ≥ cyc ∑ ( ab − 1 ) 2 − ( cyc ∑ ab − 1 ) 2 = − 2 ( cyc ∑ a b 2 c ) − 4 ab c d + 3 = 3 − 4 ab c d − 2 ( ab + c d ) ( a d + b c ) = 3 − 4 ab c d − 2 ( ab + c d ) ( A − ab − c d ) = 3 − 4 ab c d + 2 ( ab + c d − 2 A ) 2 − 2 A 2 ≥ 3 − 4 ab c d − ( a c − 2 b d − 1 ) 2 = 4 − ( a c − 1 ) 2 − ( 2 b d + 1 ) 2 ( a b − 1 ) 2 + ( b c − 1 ) 2 + ( c d − 1 ) 2 + ( d a − 1 ) 2 + ( a c − 1 ) 2 + ( 2 b d + 1 ) 2 ≥ 4.
(ab - 1)^2 + (bc - 1)^2 + (cd - 1)^2 + (da - 1)^2 + (ac - 1)^2 + (2bd + 1)^2 \geq 4.
( ab − 1 ) 2 + ( b c − 1 ) 2 + ( c d − 1 ) 2 + ( d a − 1 ) 2 + ( a c − 1 ) 2 + ( 2 b d + 1 ) 2 ≥ 4.
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