The numbers look good

Algebra Level 4

Let a a , b b and c c be non-negative real numbers such that a + b + c = 1 a + b + c = 1 . Find the maximum value of 10101 ( a b + b c + c a ) 12987 a b c . 10101(ab + bc + ca) - 12987abc.


The answer is 2886.00.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Steven Jim
Apr 18, 2017

10101 ( a b + b c + c a ) 12987 a b c = 1443 ( 7 ( a b + b c + c a ) 9 a b c ) 10101(ab + bc + ca) - 12987abc = 1443(7(ab+ bc + ca) - 9abc)

We can see that:

7 ( a + b + c ) ( a b + b c + c a ) 9 a b c 7(a + b + c)(ab + bc + ca) - 9abc

= 7 a 2 b + 7 a 2 c + 7 b 2 c + 7 b 2 a + 7 c 2 a + 7 c 2 b + 21 a b c 9 a b c =7a^{2}b + 7a^{2}c + 7b^{2}c + 7b^{2}a + 7c^{2}a + 7c^{2}b + 21abc - 9abc

= 7 a 2 b + 7 a 2 c + 7 b 2 c + 7 b 2 a + 7 c 2 a + 7 c 2 b + 12 a b c =7a^{2}b + 7a^{2}c + 7b^{2}c + 7b^{2}a + 7c^{2}a + 7c^{2}b + 12abc

= 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 =ab(7a + 7b + 4c) + bc (7b + 7c + 4a) + ca (7a + 7c + 4b) + 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 =ab(6 + a + b - 2c) + bc (6 + b + c - 2a) + ca (6 + a + c - 2b) - 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 =6ab + 6bc + 6ca + a^{2}b + a^{2}c + b^{2}c + b^{2}a + c^{2}a + c^{2}b - 6abc - 2a^{2} - 2b^{2} - 2c^{2} - 4ab - 4bc - 4ca + 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 =2ab + 2bc + 2ca + a^{2}b + a^{2}c + b^{2}c + b^{2}a + c^{2}a + c^{2}b - 6abc - 2a^{2} - 2b^{2} - 2c^{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 =2ab + 2bc + 2ca + a^{2}b + a^{2}c + b^{2}c + b^{2}a + c^{2}a + c^{2}b - 6abc - 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 =2ab + 2bc + 2ca + a^{2}b + a^{2}c + b^{2}c + b^{2}a + c^{2}a + c^{2}b - 6abc - 2a^{3} - 2b^{3} - 2c^{3} - 2a^{2}b - 2a^{2}c - 2b^{2}c - 2b^{2}a - 2c^{2}a - 2c^{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 =2ab + 2bc + 2ca - 2a^{3} - 2b^{3} - 2c^{3} - a^{2}b - a^{2}c - b^{2}c - b^{2}a - c^{2}a - c^{2}b + 2 - 6abc

= 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 + b + c)(ab + bc + ca) - 2a^{3} - 2b^{3} - 2c^{3} - a^{2}b - a^{2}c - b^{2}c - b^{2}a - c^{2}a - c^{2}b + 2 - 6abc

= 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 =2a^{2}b + 2a^{2}c + 2b^{2}c + 2b^{2}a + 2c^{2}a + 2c^{2}b + 6abc - 2a^{3} - 2b^{3} - 2c^{3} - a^{2}b - a^{2}c - b^{2}c - b^{2}a - c^{2}a - c^{2}b + 2 - 6abc

= 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^{2}b + a^{2}c + b^{2}c + b^{2}a + c^{2}a + c^{2}b - 2a^{3} - 2b^{3} - 2c^{3} + 2

= ( a b ) ( a 2 b 2 ) ( b c ) ( b 2 c 2 ) ( c a ) ( c 2 a 2 ) + 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 =- (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 min((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 => max(- (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 => max(7(a + b + c)(ab + bc + ca) - 9abc) = 2

= > m a x ( 10101 ( a b + b c + c a ) 12987 a b c ) = 1443 m a x ( 7 ( a + b + c ) ( a b + b c + c a ) 9 a b c ) = 2 1443 = 2886.00 => max(10101(ab + bc + ca) - 12987abc) = 1443max(7(a + b + c)(ab + bc + ca) - 9abc) = 2*1443 = 2886.00

Please give me a comment about this solution, as I haven't written many solutions using L a T e X LaTeX . Thanks!

Pi Han Goh
May 21, 2017

Let S n S_n denote the n th n^\text{th} symmetric sum of a , b , c a,b,c . That is, S 1 = a + b + c = 1 , S 2 = a b + a c + b c , S 3 = a b c S_1 = a+b+c= 1 , S_2 = ab+ac+bc, S_3 = abc .

We want to maximize f ( a , b , c ) : = 1443 ( 7 S 2 9 S 3 ) f(a,b,c) := 1443(7S_2 - 9S_3) . Let g ( x ) = x 3 S 1 x 2 + S 2 x S 3 g(x) = x^3 - S_1 x^2 + S_2 x - S_3 be a polynomial with roots a , b , c a,b,c .

Since a , b , c a,b,c are non-negative real numbers, then the discriminant of g ( x ) g(x) is non-negative,

S 2 2 4 S 2 3 + 4 S 3 27 S 3 2 + 18 S 2 S 3 0. S_2^2 - 4S_2^3 + 4S_3 - 27 S_3^2+ 18S_2S_3 \geq 0 .

Solving this quadratic inequality in S 3 S_3 gives

1 27 ( 2 9 S 2 2 ( 3 S 2 1 ) 3 ) S 3 1 27 ( 2 9 S 2 + 2 ( 3 S 2 1 ) 3 ) . \dfrac1{27} \left( 2 - 9S_2 - 2\sqrt{-(3S_2-1)^3} \right) \leq S_3 \leq \dfrac1{27} \left( 2 - 9S_2 + 2\sqrt{-(3S_2-1)^3} \right) .

For S 3 S_3 to be real, the expression under the radical must be non-negative, S 2 1 3 S_2 \leq \dfrac13 . So S 3 S_3 can take a minimum the value of 1 27 \dfrac1{27} when S 2 = 1 3 S_2 = \dfrac13 .

Thus, max ( f ( a , b , c ) ) = 1443 ( 7 ( max ( S 2 ) ) 1443 ( 9 ) ( min ( S 3 ) ) = 2886 \max (f(a,b,c)) = 1443( 7(\max (S_2)) - 1443(9)(\min (S_3)) =\boxed{ 2886} . This occurs when ( S 1 , S 2 , S 3 ) = ( 1 , 1 3 , 1 27 ) ( S_1, S_2, S_3) = \left( 1 , \dfrac13, \dfrac1{27} \right) , or equivalently a = b = c = 1 3 a=b=c =\dfrac13 .

Nice solution!

Steven Jim - 4 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...