There must be a pattern here!

Algebra Level 4

[ ( x + y ) ( x + z ) + ( y + z ) ( 2 y ) ] + [ ( x + 2 y ) ( x + y + z ) + y z ] \left[ \sqrt{(x+y)(x+z)} + \sqrt{(y+z)(2y)} \right] + \left[ \sqrt{(x+2y)(x+y+z)} + \sqrt{yz} \right]

Let x , y , x, \space y, and z z be real positive number, such that x + 2 y + z = 9 x + 2y + z = 9 . Find the maximum value of the expression above.

Hint: Two inequalities are summed up.

For more problem on finding maximum and minimum value, click here


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The answer is 18.

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2 solutions

Fidel Simanjuntak
Jun 11, 2017

With Cauchy-Schwarz Inequality, given a c + b d ( a + b ) ( c + d ) \sqrt{ac} + \sqrt{bd} \leq \sqrt{(a + b)(c + d)} .

First, let a = x + y ; b = y + z ; c = x + z ; d = 2 y \color{#D61F06} a = x + y \color{#333333}; \space \color{#3D99F6} b = y + z \color{#333333}; \space \color{#20A900} c = x + z\color{#333333}; \space \color{#EC7300} d = 2y

Then, we have

( x + 2 y + z ) ( x + 2 y + z ) ( x + y ) ( x + z ) + ( y + z ) ( 2 y ) 9 ( x + y ) ( x + z ) + ( y + z ) ( 2 y ) . . . ( 1 ) \begin{aligned} \sqrt{(x+2y+z)(x+2y+z)} & \geq \sqrt{\color{#D61F06}(x+y)\color{#20A900}(x+z)} + \sqrt{\color{#3D99F6}(y+z)\color{#EC7300}(2y)} \\ 9 & \geq \sqrt{(x+y)(x+z)} + \sqrt{(y+z)(2y)} \space ...(1) \end{aligned}

Second, let a = x + 2 y ; b = z ; c = x + y + z ; d = y \color{#D61F06} a = x + 2y \color{#333333}; \space \color{#3D99F6} b = z \color{#333333}; \space \color{#20A900} c = x + y + z\color{#333333}; \space \color{#EC7300} d = y

Then, we have

( x + 2 y + z ) ( x + 2 y + z ) ( x + 2 y ) ( x + y + z ) + ( z ) ( y ) 9 ( x + 2 y ) ( x + y + z ) + y z . . . ( 2 ) \begin{aligned} \sqrt{(x+2y+z)(x+2y+z)} & \geq \sqrt{\color{#D61F06}(x+2y)\color{#20A900}(x+y+z)} + \sqrt{\color{#3D99F6}(z)\color{#EC7300}(y)} \\ 9 & \geq \sqrt{(x+2y)(x+y+z)} + \sqrt{yz} \space ...(2) \end{aligned}

( 1 ) + ( 2 ) (1) + (2) gives

[ ( x + y ) ( x + z ) + ( y + z ) ( 2 y ) ] + [ ( x + 2 y ) ( x + y + z ) + y z ] 18 \left[ \sqrt{(x+y)(x+z)} + \sqrt{(y+z)(2y)} \right] + \left[ \sqrt{(x+2y)(x+y+z)} + \sqrt{yz} \right] \leq 18

Hence, the answer is 18 \boxed{18}

@Fidel Simanjuntak Good solution! Anyways, it should be "There must be a pattern here!".

Steven Jim - 3 years, 12 months ago

By the way, I see a few typing errors. Can you please fix them?

Steven Jim - 3 years, 12 months ago

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Edited. Thanks for pointing it out! What do you think about this set of problems?

Fidel Simanjuntak - 3 years, 12 months ago

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A good one, not 2 hard but not 2 easy. That's what I expect in such a set.

By the way, it's "Hence, the answer..."

Steven Jim - 3 years, 12 months ago

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@Steven Jim Oh my goodness .. Hahahahaaa,, thanks for pointing it out..

Fidel Simanjuntak - 3 years, 12 months ago

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@Fidel Simanjuntak It's okay :)

Steven Jim - 3 years, 12 months ago

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@Steven Jim What problem do you like at most?

Fidel Simanjuntak - 3 years, 12 months ago

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@Fidel Simanjuntak "K gives a great inequality" is a good one.

Steven Jim - 3 years, 12 months ago

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@Steven Jim Wew.. I like it too..

Fidel Simanjuntak - 3 years, 12 months ago

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@Fidel Simanjuntak Yeah. And I feel like others are somewhat easy to guess.

Steven Jim - 3 years, 12 months ago

@Fidel Simanjuntak By the way, try this set.

Steven Jim - 3 years, 12 months ago

Plotting the graph, taking z=9-x-2y, and vari\ying x=1 to 10, after which y goes to <0. The max of curve stay near 18 for many values.

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