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Algebra Level 5

Given that a , b a,b and c c are real numbers satisfying a + b + c = 0 a+b+c=0 and a 2 + b 2 + c 2 = 1 a^2+b^2+c^2=1 . If the maximum value of a b c abc is equal to 1 k \dfrac1k , find k 2 k^2 .


The answer is 54.

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3 solutions

Akshay Sharma
Feb 10, 2016

Since a + b + c = 0 a+b+c=0 = > => ( a + b ) 2 = ( c ) 2 (a+b)^2=(-c)^2 ,Hence a 2 + b 2 + 2 a b = c 2 a^2+b^2+2 ab=c^2 . Substituting a 2 + b 2 = 1 c 2 a^2+b^2=1-c^2 ;we get a b = c 2 1 2 ab=c^2-\frac{1}{2} = > => a b c = c 3 c 2 abc=c^3-\frac{c}{2} . Now consider a function f ( c ) = c 3 c 2 f(c)=c^3-\frac{c}{2} ,since maxima occurs when d y d x = 0 \frac{dy}{dx}=0 .Hence maxima will occur at c = 1 6 c=\frac{1}{\sqrt{6}} .Substituting value of c c [= 1 6 \frac{1}{\sqrt{6}} ] in the f ( c ) f(c) ,we get a b c = 1 3 6 abc=\frac{1}{3\sqrt{6}} .Hence k = 1 3 6 k=\frac{1}{3\sqrt{6}} = > => k 2 = 54 k^2=54

Oh, there goes gravity!(+1)

Parth Lohomi - 5 years ago
P C
May 20, 2016

Without losing generality, we assume a b 0 ab\geq 0

From the condition, we have a + b = c a+b=-c a 2 + b 2 + ( a + b ) 2 = 1 ( ) \Rightarrow a^2+b^2+(a+b)^2=1 \ (*) By Power - Mean we get a 2 + b 2 ( a + b ) 2 2 a^2+b^2\geq\frac{(a+b)^2}{2} 3 2 ( a + b ) 2 1 \Rightarrow \frac{3}{2}(a+b)^2\leq 1 c 2 2 3 c 2 6 \Leftrightarrow c^2\leq\frac{2}{3}\Rightarrow c\leq\frac{2}{\sqrt{6}} From ( ) (*) , we get 2 ( a 2 + b 2 + a b ) = 1 6 a b a b 1 6 2(a^2+b^2+ab)=1\geq 6ab\Rightarrow ab\leq\frac{1}{6} a b c 1 3 6 k = 3 6 k 2 = 54 \therefore abc\leq\frac{1}{3\sqrt{6}}\Rightarrow k=3\sqrt{6}\Rightarrow k^2=54 The equality holds when ( a ; b ; c ) = ( 1 6 ; 1 6 ; 2 6 ) (a;b;c)=\bigg(\frac{-1}{\sqrt{6}};\frac{-1}{\sqrt{6}};\frac{2}{\sqrt{6}}\bigg)

James Wilson
Jan 8, 2021

I just substituted c = a b c=-a-b in the other equations and went straight for Lagrange Multipliers: a b c = a b ( a + b ) , abc=-ab(a+b), a 2 + a b + b 2 = 1 2 . a^2+ab+b^2 = \frac{1}{2}. Applying the Lagrange Multipliers approach leads to: 2 a b b 2 = λ ( 2 a + b ) , -2ab-b^2=\lambda (2a+b), a 2 2 a b = λ ( 2 b + a ) , -a^2-2ab=\lambda (2b+a), a 2 + a b + b 2 = 1 2 . a^2+ab+b^2=\frac{1}{2}. 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 ) (-2ab-b^2)\lambda (2b+a)=\lambda (2a+b)(-a^2-2ab) b ( 2 a + b ) ( 2 b + a ) = a ( a + 2 b ) ( 2 a + b ) . \Rightarrow b(2a+b)(2b+a)=a(a+2b)(2a+b). This implies a = b a=b or b = 2 a b=-2a or a = 2 b a=-2b .

Substitute these, in turn, into a 2 + a b + b 2 = 1 2 . a^2+ab+b^2=\frac{1}{2}.

a = b a = ± 1 6 , b = ± 1 6 a=b\Rightarrow a=\pm\frac{1}{\sqrt{6}}, b=\pm\frac{1}{\sqrt{6}}

a b c = 1 54 . \Rightarrow abc=\mp\frac{1}{\sqrt{54}}.

b = 2 a a = ± 1 6 , b = 2 6 b=-2a \Rightarrow a=\pm\frac{1}{\sqrt{6}}, b=\mp\frac{2}{\sqrt{6}}

a b c = 1 54 . \Rightarrow abc=\mp\frac{1}{\sqrt{54}}.

a = 2 b a = ± 2 6 , b = 1 6 a=-2b \Rightarrow a=\pm\frac{2}{\sqrt{6}}, b=\mp\frac{1}{\sqrt{6}}

a b c = ± 1 54 . \Rightarrow abc = \pm\frac{1}{\sqrt{54}}.

Because a a and b b are confined to the ellipse a 2 + a b + b 2 = 1 2 a^2+ab+b^2=\frac{1}{2} , I believe we can conclude that the global maximum is 1 54 \frac{1}{\sqrt{54}} . (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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