[ ( x + y ) ( x + z ) + ( y + z ) ( 2 y ) ] + [ ( x + 2 y ) ( x + y + z ) + y z ]
Let x , y , and z be real positive number, such that x + 2 y + z = 9 . Find the maximum value of the expression above.
Hint: Two inequalities are summed up.
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@Fidel Simanjuntak Good solution! Anyways, it should be "There must be a pattern here!".
By the way, I see a few typing errors. Can you please fix them?
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Edited. Thanks for pointing it out! What do you think about this set of problems?
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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..."
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@Steven Jim – Oh my goodness .. Hahahahaaa,, thanks for pointing it out..
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@Fidel Simanjuntak – It's okay :)
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@Steven Jim – What problem do you like at most?
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@Fidel Simanjuntak – "K gives a great inequality" is a good one.
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@Steven Jim – Wew.. I like it too..
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@Fidel Simanjuntak – Yeah. And I feel like others are somewhat easy to guess.
@Fidel Simanjuntak – By the way, try this set.
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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With Cauchy-Schwarz Inequality, given a c + b d ≤ ( a + b ) ( c + d ) .
First, let a = x + y ; b = y + z ; c = x + z ; d = 2 y
Then, we have
( x + 2 y + z ) ( x + 2 y + z ) 9 ≥ ( x + y ) ( x + z ) + ( y + z ) ( 2 y ) ≥ ( x + y ) ( x + z ) + ( y + z ) ( 2 y ) . . . ( 1 )
Second, let a = x + 2 y ; b = z ; c = x + y + z ; d = y
Then, we have
( x + 2 y + z ) ( x + 2 y + z ) 9 ≥ ( x + 2 y ) ( x + y + z ) + ( z ) ( y ) ≥ ( x + 2 y ) ( x + y + z ) + y z . . . ( 2 )
( 1 ) + ( 2 ) gives
[ ( x + y ) ( x + z ) + ( y + z ) ( 2 y ) ] + [ ( x + 2 y ) ( x + y + z ) + y z ] ≤ 1 8
Hence, the answer is 1 8