a 2 + b 2 + c 2 + d 2 + 1 2 < a b + 3 b + 6 c + 2 d
How many sets of 4 positive integers ( a , b , c , d ) that satisfy the inequality above?
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Since a , b , c , d are all positive integers, we will find ( a , b , c , d ) such that
a 2 + b 2 + c 2 + d 2 + 1 3 ≤ a b + 3 b + 6 c + 2 d ⟺ a 2 + b 2 + c 2 + d 2 + 1 3 − a b + 3 b + 6 c + 2 d ≤ 0 ⟺ ( a − 2 1 b ) 2 + 3 ( 2 1 b − 1 ) 2 + ( c − 3 ) 2 + ( d − 1 ) 2 ≤ 0 ⟺ a = 1 , b = 2 , c = 3 , d = 1 .
Hence, there is 1 set ( a , b , c , d ) .
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a 2 + b 2 + c 2 + d 2 + 1 2 a 2 − a b + b 2 − 3 b + c 2 − 6 c + d 2 − 2 d + 1 2 a 2 − a b + 4 b 2 + 4 3 ( b 2 − 4 b + 4 ) + c 2 − 6 c + 9 + d 2 − 2 d + 1 − 1 ( a − 2 b ) 2 + 4 3 ( b − 2 ) 2 + ( c − 3 ) 2 + ( d − 1 ) 2 < a b + 3 b + 6 c + 2 d < 0 < 0 < 1
We note that the LHS ≤ 0 and equality occurs when a = 1 , b = 2 , c = 3 and d = 1 . Since the LHS is integral, other sets of ( a , b , c , d ) will make it ≥ 1 . Therefore, there is only 1 solution.