Let a , b and c be non-negative real numbers such that a + b + c = 1 . Find the maximum value of 1 0 1 0 1 ( a b + b c + c a ) − 1 2 9 8 7 a b c .
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Let S n denote the n th symmetric sum of a , b , c . That is, S 1 = a + b + c = 1 , S 2 = a b + a c + b c , S 3 = a b c .
We want to maximize f ( a , b , c ) : = 1 4 4 3 ( 7 S 2 − 9 S 3 ) . Let g ( x ) = x 3 − S 1 x 2 + S 2 x − S 3 be a polynomial with roots a , b , c .
Since a , b , c are non-negative real numbers, then the discriminant of g ( x ) is non-negative,
S 2 2 − 4 S 2 3 + 4 S 3 − 2 7 S 3 2 + 1 8 S 2 S 3 ≥ 0 .
Solving this quadratic inequality in S 3 gives
2 7 1 ( 2 − 9 S 2 − 2 − ( 3 S 2 − 1 ) 3 ) ≤ S 3 ≤ 2 7 1 ( 2 − 9 S 2 + 2 − ( 3 S 2 − 1 ) 3 ) .
For S 3 to be real, the expression under the radical must be non-negative, S 2 ≤ 3 1 . So S 3 can take a minimum the value of 2 7 1 when S 2 = 3 1 .
Thus, max ( f ( a , b , c ) ) = 1 4 4 3 ( 7 ( max ( S 2 ) ) − 1 4 4 3 ( 9 ) ( min ( S 3 ) ) = 2 8 8 6 . This occurs when ( S 1 , S 2 , S 3 ) = ( 1 , 3 1 , 2 7 1 ) , or equivalently a = b = c = 3 1 .
Nice solution!
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1 0 1 0 1 ( a b + b c + c a ) − 1 2 9 8 7 a b c = 1 4 4 3 ( 7 ( a b + b c + c a ) − 9 a b c )
We can see that:
7 ( a + b + c ) ( a b + b c + c a ) − 9 a b c
= 7 a 2 b + 7 a 2 c + 7 b 2 c + 7 b 2 a + 7 c 2 a + 7 c 2 b + 2 1 a b c − 9 a b c
= 7 a 2 b + 7 a 2 c + 7 b 2 c + 7 b 2 a + 7 c 2 a + 7 c 2 b + 1 2 a b c
= a b ( 7 a + 7 b + 4 c ) + b c ( 7 b + 7 c + 4 a ) + c a ( 7 a + 7 c + 4 b ) + 2 − 2
= a b ( 6 + a + b − 2 c ) + b c ( 6 + b + c − 2 a ) + c a ( 6 + a + c − 2 b ) − 2 ( a + b + c ) 2 + 2
= 6 a b + 6 b c + 6 c a + a 2 b + a 2 c + b 2 c + b 2 a + c 2 a + c 2 b − 6 a b c − 2 a 2 − 2 b 2 − 2 c 2 − 4 a b − 4 b c − 4 c a + 2
= 2 a b + 2 b c + 2 c a + a 2 b + a 2 c + b 2 c + b 2 a + c 2 a + c 2 b − 6 a b c − 2 a 2 − 2 b 2 − 2 c 2 + 2
= 2 a b + 2 b c + 2 c a + a 2 b + a 2 c + b 2 c + b 2 a + c 2 a + c 2 b − 6 a b c − 2 ( a + b + c ) ( a 2 + b 2 + c 2 ) + 2
= 2 a b + 2 b c + 2 c a + a 2 b + a 2 c + b 2 c + b 2 a + c 2 a + c 2 b − 6 a b c − 2 a 3 − 2 b 3 − 2 c 3 − 2 a 2 b − 2 a 2 c − 2 b 2 c − 2 b 2 a − 2 c 2 a − 2 c 2 b + 2
= 2 a b + 2 b c + 2 c a − 2 a 3 − 2 b 3 − 2 c 3 − a 2 b − a 2 c − b 2 c − b 2 a − c 2 a − c 2 b + 2 − 6 a b c
= 2 ( a + b + c ) ( a b + b c + c a ) − 2 a 3 − 2 b 3 − 2 c 3 − a 2 b − a 2 c − b 2 c − b 2 a − c 2 a − c 2 b + 2 − 6 a b c
= 2 a 2 b + 2 a 2 c + 2 b 2 c + 2 b 2 a + 2 c 2 a + 2 c 2 b + 6 a b c − 2 a 3 − 2 b 3 − 2 c 3 − a 2 b − a 2 c − b 2 c − b 2 a − c 2 a − c 2 b + 2 − 6 a b c
= a 2 b + a 2 c + b 2 c + b 2 a + c 2 a + c 2 b − 2 a 3 − 2 b 3 − 2 c 3 + 2
= − ( a − b ) ( a 2 − b 2 ) − ( b − c ) ( b 2 − c 2 ) − ( c − a ) ( c 2 − a 2 ) + 2
= − ( a − b ) 2 ( a + b ) − ( b − c ) 2 ( b + c ) − ( c − a ) 2 ( c + a ) + 2
As a,b,c are non negative reals, we can easily see that
m i n ( ( a − b ) 2 ( a + b ) + ( b − c ) 2 ( b + c ) + ( c − a ) 2 ( c + a ) − 2 ) = − 2
= > m a x ( − ( a − b ) 2 ( a + b ) − ( b − c ) 2 ( b + c ) − ( c − a ) 2 ( c + a ) + 2 ) = 2
= > m a x ( 7 ( a + b + c ) ( a b + b c + c a ) − 9 a b c ) = 2
= > m a x ( 1 0 1 0 1 ( a b + b c + c a ) − 1 2 9 8 7 a b c ) = 1 4 4 3 m a x ( 7 ( a + b + c ) ( a b + b c + c a ) − 9 a b c ) = 2 ∗ 1 4 4 3 = 2 8 8 6 . 0 0
Please give me a comment about this solution, as I haven't written many solutions using L a T e X . Thanks!