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Let x,y, and z be real numbers such that x + y + z = 6 and $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$$ = 2. Find $$\frac{x + y}{z} + \frac{y + z}{x} + \frac{x + z}{y}.$$

Sep 15, 2019

#1
+6196
+1

$$\dfrac 1 x + \dfrac 1 y + \dfrac 1 z = 2\\ \dfrac{x+y+z}{x+y+z}\left( \dfrac 1 x + \dfrac 1 y + \dfrac 1 z\right) = 2\\ \dfrac 1 6 \left(1+\dfrac{y+z}{x} + 1 + \dfrac{x+z}{y}+1+\dfrac{x+y}{z}\right) = 2\\ 3 + \left(\dfrac{y+z}{x} + \dfrac{x+z}{y}+\dfrac{x+y}{z}\right) = 12\\ \left(\dfrac{y+z}{x} + \dfrac{x+z}{y}+\dfrac{x+y}{z}\right) = 9$$

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Sep 15, 2019
#2
+2856
+1

Good job, Rom!

CalculatorUser  Sep 15, 2019
#3
+111433
+1

x + y + z  = 6       (1)

1/x + 1/y + 1/z = 2     ⇒  [ yz +  xz + xy] / xyz  = 2  ⇒  [ xy + xz + yz]  = 2xyz      (2)

x + y          y + z            x + z

Find        ____  +   ______   +  ______    =

z               x                   y

xy ( x + y)   + yz( y + z)  +  xz ( x + z)

________________________________  =

xyz

xy  ( 6 - z)   + yz ( 6 - x)   +  xz ( 6 -y)

______________________________      =

xyz

6xy - xyz  + 6yz - xyz  +  6xz - xyz

_______________________________  =

xyz

6 [ xy  + xz +  yz ] -  3 [ xxz ]

_______________________   =

xyz

6  [ 2xyz ]   - 3 [xyz]

________________   =

xyz

xyz [ 6*2  - 3 ]

____________  =

xyz

6*2  - 3  =

12 - 3  =

9

Sep 15, 2019