An algebra problem by Priyanshu Mishra

Algebra Level 5

If ( x , y , z ) (x,y,z) are non-negative reals and x y + y z + z x = 1 xy + yz + zx = 1 ,

The maximum value of the expression below is of the form a b c \large \frac { a }{ b\sqrt { c } } .

x ( 1 y 2 ) ( 1 z 2 ) + y ( 1 z 2 ) ( 1 x 2 ) + z ( 1 x 2 ) ( 1 y 2 ) \large\ x\left( 1-{ y }^{ 2 } \right) \left( 1-{ z }^{ 2 } \right) + y\left( 1-{ z }^{ 2 } \right) \left( 1-{ x }^{ 2 } \right) + z\left( 1-{ x }^{ 2 } \right) \left( 1-{ y }^{ 2 } \right)

a , b , c a,b,c are all positive integers with a , b a,b coprime and c c square-free.

Find a + b + c a + b + c .


The answer is 10.

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

Rajen Kapur
Feb 3, 2017

Given expression: c y l x ( 1 ( y 2 + z 2 ) + y 2 z 2 ) = x c y l ( x y 2 ) + x y z ( x y + y z + z x ) = 4 x y z \sum_{cyl} x(1-(y^2+z^2)+y^2z^2)=\sum x-\sum_{cyl}(xy^2)+xyz(xy+yz+zx)=4xyz Using x = ( x + y + z ) ( x y + y z + z x ) = ( x y 2 ) + 3 x y z \sum x=(x+y+z)(xy+yz+zx)=\sum (xy^2)+3xyz Now A.M. - G.M.: x y + y z + z x 3 ( x y z ) 3 2 \frac {xy+yz+zx}{3}\geq(xyz)^\frac {3}{2} We get a = 4 , b = 3 , c = 3 a=4, b=3, c=3 , hence answer 10.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...