What sort of inequalities would help with this?
Let be real nonzero numbers such that and Compute the largest possible value of .
The answer is 56.
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1 solution
From the second condition, we have 2 a b c = 2 a b + 2 b c + 2 c a + 2 . Thus, 2 ( a b + b c + c a ) = ( a + b + c ) 2 − a 2 − b 2 − c 2 = 1 4 4 − a 2 − b 2 − c 2 .
Therefore, 2 a b c = 1 4 6 − a 2 − b 2 − c 2 . We now have 2 a b c − 2 ( a + 2 b − 3 c ) = 1 4 6 − ( a 2 + 2 a ) − ( b 2 + 4 b ) − ( c 2 − 6 c ) = 1 6 0 − ( ( a + 1 ) 2 + ( b + 2 ) 2 + ( c − 3 ) 2 ) .
By QM-AM, ( a + 1 ) 2 + ( b + 2 ) 2 + ( c − 3 ) 2 ≥ 3 ( 3 ( a + 1 ) + ( b + 2 ) + ( c − 3 ) ) 2 = 4 8 . Therefore, 2 a b c − 2 ( a + 2 b − 3 c ) ≤ 1 6 0 − 2 × 5 6 = 1 1 2 , so the maximum value of a b c − ( a + 2 b − 3 c ) = 5 6 . This occurs at a = 3 , b = 2 , c = 7 .