Mensa Intermediate Algebra Inequality Problem Solution

 

The Problem:

Given xyz = 1 (x, y and z positive real numbers) prove the following inequality:

Solution:

Multiply the left side by xyz equal to 1);

yz/ x² (y + z)  +  xz/ y² (x + z)  +   xy/ z² (x + y) 

Multiply the left side by xyz again:

y²  z²/ (xy + xz)  +   x²  z²/ (yx +yz)  +    x²  y²/ (zx + zy) 

Apply Titu's Lemma:

y²  z²/ (xy + xz)  +   x²  z²/ (yx +yz)  +    x²  y²/ (zx + zy) >

(yz + xz + xy)² /(xy +xz + yx + yz + zx + zy)

The right side can now be simplified:

(yz + xz + xy)² /(xy +xz + yx + yz + zx + zy) =

(yz + xz + xy)² / 2 (yz + xz + xy) =

(yz + xz + xy)/ 2   

Then:

 

y²  z²/ (xy + xz) +   x²  z²/ (yx +yz)  +    x²  y²/ (zx + zy)  >

(yz + xz + xy)/ 2   

Now apply the AM-GM inequality to the right -side numerator

(yz + xz + xy):

  

(yz + xz + xy)/3 >  

 ³Ö(yz · xz  ·  xy) =    ³Ö(x² y² z²) =   1   

Multiply both sides by 3:

(yz + xz + xy)  > 3

y²  z²/ (xy + xz) +   x²  z²/ (yx +yz)  +    x²  y²/ (zx + zy)  >

(yz + xz + xy)/2  > 3/2

1/ (x³ y + x³ z)  +   1/ (y³ x +y³ z)  +   1/ (z³x + z³ y) > 3/2

Q.E.D.