Consequence #1

Algebra Level pending

Let x , y , z x,y,z be the positive real numbers such that x y z = 1 xyz=1 . Find the maximum value of A = 1 ( x + 1 ) 2 + y 2 + 1 + 1 ( y + 1 ) 2 + z 2 + 1 + 1 ( z + 1 ) 2 + x 2 + 1 A=\frac { 1 }{ { \left( x+1 \right) }^{ 2 }+{ y }^{ 2 }+1 } +\frac { 1 }{ { \left( y+1 \right) }^{ 2 }+{ z }^{ 2 }+1 } +\frac { 1 }{ { \left( z+1 \right) }^{ 2 }+x^{ 2 }+1 }

3 4 \frac { 3 }{ 4 } 1 2 \frac { 1 }{ 2 } 1 3 \frac { 1 }{ 3 } 2

Linkin Duck by Linkin Duck , 33, Vietnam

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

Linkin Duck
Mar 24, 2017

We notice that:

( x + 1 ) 2 + y 2 + 1 = x 2 + y 2 + 2 x + 2 = ( x − y ) 2 + 2 ( x y + x + 1 ) ≥ 2 ( x y + x + 1 ) { \left( x+1 \right) }^{ 2 }+{ y }^{ 2 }+1={ x }^{ 2 }+{ y }^{ 2 }+2x+2={ \left( x-y \right) }^{ 2 }+2\left( xy+x+1 \right) \ge 2\left( xy+x+1 \right)

and x , y > 0 x,y>0

⟹ 1 ( x + 1 ) 2 + y 2 + 1 ≤ 1 2 ( x y + x + 1 ) \Longrightarrow \frac { 1 }{ { \left( x+1 \right) }^{ 2 }+{ y }^{ 2 }+1 } \le \frac { 1 }{ 2\left( xy+x+1 \right) }

Similarly, 1 ( y + 1 ) 2 + z 2 + 1 ≤ 1 2 ( y z + y + 1 ) , 1 ( z + 1 ) 2 + x 2 + 1 ≤ 1 2 ( z x + z + 1 ) . \frac { 1 }{ { \left( y+1 \right) }^{ 2 }+{ z }^{ 2 }+1 } \le \frac { 1 }{ 2\left( yz+y+1 \right) } ,\\ \frac { 1 }{ { \left( z+1 \right) }^{ 2 }+{ x }^{ 2 }+1 } \le \frac { 1 }{ 2\left( zx+z+1 \right) }.

So, A ≤ 1 2 ( 1 x y + x + 1 + 1 y z + y + 1 + 1 z x + z + 1 ) = 1 2 A\le \frac { 1 }{ 2 } \left( \frac { 1 }{ xy+x+1 } +\frac { 1 }{ yz+y+1 } +\frac { 1 }{ zx+z+1 } \right) =\frac { 1 }{ 2 }

(by applying the result of this problem )

When x = y = z = 1 x=y=z=1 we have A = 1 / 2 A=1/2 , so A m i n = 1 2 { A }_{ min }=\frac { 1 }{ 2 }

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...