Given that a , b and c are real numbers satisfying a + b + c = 0 and a 2 + b 2 + c 2 = 1 . If the maximum value of a b c is equal to k 1 , find k 2 .
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Without losing generality, we assume a b ≥ 0
From the condition, we have a + b = − c ⇒ a 2 + b 2 + ( a + b ) 2 = 1 ( ∗ ) By Power - Mean we get a 2 + b 2 ≥ 2 ( a + b ) 2 ⇒ 2 3 ( a + b ) 2 ≤ 1 ⇔ c 2 ≤ 3 2 ⇒ c ≤ 6 2 From ( ∗ ) , we get 2 ( a 2 + b 2 + a b ) = 1 ≥ 6 a b ⇒ a b ≤ 6 1 ∴ a b c ≤ 3 6 1 ⇒ k = 3 6 ⇒ k 2 = 5 4 The equality holds when ( a ; b ; c ) = ( 6 − 1 ; 6 − 1 ; 6 2 )
I just substituted c = − a − b in the other equations and went straight for Lagrange Multipliers: a b c = − a b ( a + b ) , a 2 + a b + b 2 = 2 1 . Applying the Lagrange Multipliers approach leads to: − 2 a b − b 2 = λ ( 2 a + b ) , − a 2 − 2 a b = λ ( 2 b + a ) , a 2 + a b + b 2 = 2 1 . Multiplying the first two equations, I get: ( − 2 a b − b 2 ) λ ( 2 b + a ) = λ ( 2 a + b ) ( − a 2 − 2 a b ) ⇒ b ( 2 a + b ) ( 2 b + a ) = a ( a + 2 b ) ( 2 a + b ) . This implies a = b or b = − 2 a or a = − 2 b .
Substitute these, in turn, into a 2 + a b + b 2 = 2 1 .
a = b ⇒ a = ± 6 1 , b = ± 6 1
⇒ a b c = ∓ 5 4 1 .
b = − 2 a ⇒ a = ± 6 1 , b = ∓ 6 2
⇒ a b c = ∓ 5 4 1 .
a = − 2 b ⇒ a = ± 6 2 , b = ∓ 6 1
⇒ a b c = ± 5 4 1 .
Because a and b are confined to the ellipse a 2 + a b + b 2 = 2 1 , I believe we can conclude that the global maximum is 5 4 1 . (If my reasoning is incorrect, please let me know. Otherwise, please point me to the right theorem or elaborated reasoning to conclude this.)
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Since a + b + c = 0 = > ( a + b ) 2 = ( − c ) 2 ,Hence a 2 + b 2 + 2 a b = c 2 . Substituting a 2 + b 2 = 1 − c 2 ;we get a b = c 2 − 2 1 = > a b c = c 3 − 2 c . Now consider a function f ( c ) = c 3 − 2 c ,since maxima occurs when d x d y = 0 .Hence maxima will occur at c = 6 1 .Substituting value of c [= 6 1 ] in the f ( c ) ,we get a b c = 3 6 1 .Hence k = 3 6 1 = > k 2 = 5 4