Strange!

Level 2

If x , y x,y and z z are positive real numbers such that

2 x y = 13 1 − z + 1 + z 2xy = \dfrac{13}{1-z} +1+z

Find the minimum value of the expression below

x 2 + y 2 + z 2 − 2 x y z x^2+y^2+z^2-2xyz


The answer is 14.00.

A Former Brilliant Member by A Former Brilliant Member

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

The given condition can be written as : 2 x y ( 1 − z ) + z 2 = 14 2xy (1-z) + z^2 = 14

Now the expression which we need to minimize is : x 2 + y 2 + z 2 − 2 x y z x^2+y^2+z^2-2xyz x 2 + y 2 − 2 x y + 2 x y + z 2 − 2 x y z x^2+y^2-2xy+2xy+z^2-2xyz ( x − y ) 2 + 2 x y ( 1 − z ) + z 2 ≥ 2 x y ( 1 − z ) + z 2 (x-y)^2 + 2xy(1-z) +z^2 \ge 2xy(1-z) +z^2 ( x − y ) 2 + 2 x y ( 1 − z ) + z 2 ≥ 14 (x-y)^2 + 2xy(1-z) +z^2 \ge 14

Hence the answer is 14.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...