Cool Inequality #2 Promotion

Algebra Level 5

( ∑ i = 1 3 6 x 1 x 2 x 3 6 x i ) 4 \large{\left(\sum _{ i=1 }^{ 3 }{ \frac { \sqrt{6x_{1}x_{2}x_{3}} }{ 6{ x }_{ i } } }\right)^{4}}

If x 1 , x 2 , x 3 > 0 x_{1},x_{2},x_{3} > 0 such that ∑ cyclic ( x 1 × x 2 ) 2 = 6 x 1 x 2 x 3 \displaystyle\sum_{\text{cyclic} }^{ }{ ({ x }_{ 1 }\times { x }_{ 2 })^{2} } = 6x_{1}x_{2}x_{3} . Find the maximum value of the expression above.


The answer is 9.

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Department 8 by Department 8 , 27, Israel

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2 solutions

Department 8
Sep 12, 2015

By cauchy-Schwarz inequality/ Titu's Lemma, we have:-

( ( x 1 x 2 ) 2 + ( x 2 x 3 ) 2 + ( x 3 x 1 ) 2 ) ( 1 + 1 + 1 ) ≥ ( x 1 x 2 + x 2 x 3 + x 3 x 1 ) 2 3 ≥ ( x 1 x 2 + x 2 x 3 + x 3 x 1 6 x 1 x 2 x 3 ) 2 9 ≥ ( x 1 x 2 6 x 1 x 2 x 3 × 6 x 1 x 2 x 3 6 x 1 x 2 x 3 + x 2 x 3 6 x 1 x 2 x 3 × 6 x 1 x 2 x 3 6 x 1 x 2 x 3 + x 3 x 1 6 x 1 x 2 x 3 × 6 x 1 x 2 x 3 6 x 1 x 2 x 3 ) 4 9 ≥ ( ∑ i = 1 3 6 x 1 x 2 x 3 6 x i ) 4 \Large{\left( ({ { x }_{ 1 }{ x }_{ 2 } })^{ 2 }+{ ({ x }_{ 2 } }{ x }_{ 3 })^{ 2 }+{ ({ x }_{ 3 }{ x }_{ 1 }) }^{ 2 } \right) (1+1+1)\ge { \left( { x }_{ 1 }{ x }_{ 2 }+{ x }_{ 2 }{ x }_{ 3 }+{ x }_{ 3 }{ x }_{ 1 } \right) }^{ 2 }\\ 3\ge { \left( \frac { { x }_{ 1 }{ x }_{ 2 }+{ x }_{ 2 }{ x }_{ 3 }+{ x }_{ 3 }{ x }_{ 1 } }{ \sqrt { 6{ x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 } } } \right) }^{ 2 }\\ 9\ge { \left( \frac { { x }_{ 1 }{ x }_{ 2 } }{ \sqrt { 6{ x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 } } } \times \frac { \sqrt { 6{ x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 } } }{ \sqrt { 6{ x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 } } } +\frac { { x }_{ 2 }{ x }_{ 3 } }{ \sqrt { 6{ x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 } } } \times \frac { \sqrt { 6{ x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 } } }{ \sqrt { 6{ x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 } } } +\frac { { x }_{ 3 }{ x }_{ 1 } }{ \sqrt { 6{ x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 } } } \times \frac { \sqrt { 6{ x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 } } }{ \sqrt { 6{ x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 } } } \right) }^{ 4 }\\ 9\ge (\sum _{ i=1 }^{ 3 }{ \frac { \sqrt { 6{ x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 } } }{ 6{ x }_{ i } } )^{4} }}

The maximum value would be 9 \boxed{9}

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my answer was 16. I applied AM GM property.

Deepanshu Vijay - 5 years, 8 months ago

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I think you got confused with brackets around summation sign that confused my friends too.

Department 8 - 5 years, 8 months ago

What are the values of x 1 , x 2 , x 3 x_1, x_2, x_3 that give 16?

Calvin Lin Staff - 5 years, 8 months ago

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Sir that's my question, why is AM-GM not working here.

Akshat Sharma - 5 years, 8 months ago

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@Akshat Sharma – Substantiate your comment by showing your actual working, instead of just stating a very general approach. There could be many reasons why "it is not working", and that depends greatly on the way that you presented it.

In particular, why is it the maximum, and not just an upper bound? E.g. We know that x 2 ≥ 0 x^2 \geq 0 and ( x + 1 ) 2 ≥ 0 (x+1)^2 \geq 0 . Does this mean that the minimum of x 2 + ( x + 1 ) 2 x^2 + (x+1)^2 is 0?

Calvin Lin Staff - 5 years, 8 months ago

I also did it using AM-GM amd I got the answer as 9.

Samarth Agarwal - 5 years, 8 months ago

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How had you got that can you please tell.

Akshat Sharma - 5 years, 8 months ago
Chew-Seong Cheong
Oct 13, 2015

Let the expression be S S , then we have:

S = ( 6 x 1 x 2 x 3 6 [ 1 x 1 + 1 x 2 + 1 x 3 ] ) 4 = ( 6 x 1 x 2 x 3 6 x 1 x 2 x 3 ( x 1 x 2 + x 2 x 3 + x 3 x 1 ) ) 4 = ( x 1 x 2 + x 2 x 3 + x 3 x 1 ) 4 ( 6 x 1 x 2 x 3 ) 2 See Note. ≤ ( 18 x 1 x 2 x 3 ) 2 ( 6 x 1 x 2 x 3 ) 2 = 9 \begin{aligned} S & = \left( \frac{\sqrt{6x_1x_2x_3}}{6} \left[\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} \right] \right)^4 \\ & = \left( \frac{\sqrt{6x_1x_2x_3}}{6x_1x_2x_3} \left( x_1x_2 + x_2x_3 + x_3x_1 \right) \right)^4 \\ & = \frac {\color{#3D99F6}{( x_1x_2 + x_2x_3 + x_3x_1)^4}}{(6x_1x_2x_3)^2} \quad \quad \small \color{#3D99F6}{\text{See Note.}} \\ & \le \frac {\color{#3D99F6}{( 18x_1x_2 x_3)^2}}{(6x_1x_2x_3)^2} \\ & = \boxed{9} \end{aligned}

\\

Note: By Cauchy-Schwarz inequality, we have \color{#3D99F6}{\text{Note: By Cauchy-Schwarz inequality, we have }}

( x 1 x 2 + x 2 x 3 + x 3 x 1 ) 2 ≤ 3 ( ( x 1 x 2 ) 2 + ( x 2 x 3 ) 2 + ( x 3 x 1 ) 2 ) = 18 x 1 x 2 x 3 \color{#3D99F6}{( x_1x_2 + x_2x_3 + x_3x_1)^2 \le 3 \left( (x_1x_2)^2 + (x_2x_3)^2 + (x_3x_1)^2\right) = 18x_1x_2x_3}

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