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

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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 }

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