This made me won
Let x , y , and z be real numbers such that x + y + z = 6 and x 1 + y 1 + z 1 = 2 . Find the value of z ( x + y ) + x ( y + z ) + y ( x + z ) .
The answer is 9.
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4 solutions
Very nice,simple and straightforward method keep it up!
We rewrote the question to z 6 − z + x 6 − x + y 6 − y = 6 ( x 1 + y 1 + z 1 ) − 3 = 6 ( 2 ) − 3 = 9
I'm sorry about the problem being posted prior to this, I admit I was mistaken.. :D
No problem its Ok.Never delete a problem when it is wrong or reported by someone.Brilliant Staff will fix it.
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Ahh, i didn't know that.. I have already deleted the old one though..
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Ok, Never do this again. :-)
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@A Former Brilliant Member – Sure.. Thanks.. :)
The title should be
This made me win ...
To be Grammatically correct
As x + y + z = 6 So, ( x + y ) / z + ( y + z ) / x + ( z + x ) / y = ( 6 − z ) / z + ( 6 − x ) / x + ( 6 − y ) / y = 6 ( 1 / x + 1 / y + 1 / z ) − 3 = 2 ∗ 6 − 3 = 9
⇒ ( x + y + z ) × ( x 1 + y 1 + z 1 ) = 6 × 2
⇒ 3 + ( y x + z + z x + y + x y + z ) = 1 2
⇒ y x + z + z x + y + x y + z = 9