Prove Inequality: $\frac{1}{2}$ Bound w/ x,y,z>0

[Albert1]

x>0 ,y>0 ,z>0 and xyz=1 ,prove :

$ \dfrac{1}{(x+1)^2+y^2+1}+\dfrac{1}{(y+1)^2+z^2+1}+\dfrac{1}{(z+1)^2+x^2+1}\leq \dfrac{1}{2}$

 

[chisigma]

Re: Prove an inequality

x>0 ,y>0 ,z>0 and xyz=1 ,prove :

$ \dfrac{1}{(x+1)^2+y^2+1}+\dfrac{1}{(y+1)^2+z^2+1}+\dfrac{1}{(z+1)^2+x^2+1}\leq \dfrac{1}{2}$

It is quite evident the intrinsic symmetry of the expression, in the sense that x, y and z can be swapped and nothing change. Thqat suggests that it must be x=y=z=1 and in this case any memeber of the sum has value $\frac{1}{6}$ so that the sum is $\frac{1}{2}$ which is the maximum of the function with the condition x y z =1...

Kind regards

$\chi$ $\sigma$

 

[Albert1]

Re: Prove an inequality

x>0 ,y>0 ,z>0 and xyz=1 ,prove :

$ \dfrac{1}{(x+1)^2+y^2+1}+\dfrac{1}{(y+1)^2+z^2+1}

+\dfrac{1}{(z+1)^2+x^2+1}\leq \dfrac{1}{2}$

for using $xyz=1$ and $AM\geq GM$

$\dfrac {z}{z}\times \dfrac{1}{(x+1)^2+y^2+1}\leq

\dfrac{z}{2(xz+z+1)}----(1)$

$\dfrac{xz}{xz}\times \dfrac{1}{(y+1)^2+z^2+1}\leq

\dfrac{xz}{2(xz+z+1)}----(2)$

$\dfrac{1}{(z+1)^2+x^2+1}\leq \dfrac{1}{2(xz+z+1)}----(3)$

(1)+(2)+(3) the proof is done

 

[caffeinemachine]

Re: Prove an inequality

It is quite evident the intrinsic symmetry of the expression, in the sense that x, y and z can be swapped and nothing change. Thqat suggests that it must be x=y=z=1 and in this case any memeber of the sum has value $\frac{1}{6}$ so that the sum is $\frac{1}{2}$ which is the maximum of the function with the condition x y z =1...

Kind regards

$\chi$ $\sigma$

Hello Chisigma,

Your reasoning is not clear to me.
I guess you have implicitly assumed, and not proved, that a maxima exists.
Moreover, even if the existence of a maxima is settled, there may me multiple maximas.
I don't see how the symmetry is sufficient to conclude that $1/2$ is the maximum even if its given that a global maxima exists.