Comparison between x²+y²+z² and 1/x²+1/y²+1/z²

[anemone]

Let $x,\,y,\,z$ be positive real numbers such that $x+y+z=3$.

Prove that $\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\ge x^2+y^2+z^2$.

 

[Albert1]

Let $x,\,y,\,z$ be positive real numbers such that $x+y+z=3---(1)$.

Prove that $\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\ge x^2+y^2+z^2$.

my solution:

Spoiler

using Largrange multipliers :
let :$f(x,y,z,\lambda)=(\dfrac {1}{x^2}-x^2)+(\dfrac {1}{y^2}-y^2)+(\dfrac {1}{z^2}-z^2)+\lambda (x+y+z-3)$
we have:$\lambda=2x+\dfrac {2}{x^3}=2y+\dfrac {2}{y^3}=2z+\dfrac {2}{z^3}=4 ,\,\,\,for \,\,(x=y=z, \,\, and \,\,x+y+z=3)$
check another point $f(x,y,z,4)=f(1,\dfrac {4}{3},\dfrac {2}{3},4)=\dfrac {85}{144}>0=f(1,1,1,4)$
$\therefore (\dfrac {1}{x^2}-x^2)+(\dfrac {1}{y^2}-y^2)+(\dfrac {1}{z^2}-z^2)+4(x+y+z-3)\geq 0$
that is:$(\dfrac {1}{x^2}-x^2)+\dfrac {1}{y^2}-y^2)+(\dfrac {1}{z^2}-z^2)\geq 0$
and the proof is done

 

[MarkFL]

my solution:

Spoiler

using Largrange multipliers :
let :$f(x,y,z,\lambda)=(\dfrac {1}{x^2}-x^2)+(\dfrac {1}{y^2}-y^2)+(\dfrac {1}{z^2}-z^2)+\lambda (x+y+z-3)$
we have:$\lambda=2x+\dfrac {2}{x^3}=2y+\dfrac {2}{y^3}=2z+\dfrac {2}{z^3}=4 ,\,\,\,for \,\,(x=y=z, \,\, and \,\,x+y+z=3)$
check another point $f(x,y,z,4)=f(1,\dfrac {4}{3},\dfrac {2}{3},4)=\dfrac {85}{144}>0=f(1,1,1,4)$
$\therefore (\dfrac {1}{x^2}-x^2)+(\dfrac {1}{y^2}-y^2)+(\dfrac {1}{z^2}-z^2)+4(x+y+z-3)\geq 0$
that is:$(\dfrac {1}{x^2}-x^2)+\dfrac {1}{y^2}-y^2)+(\dfrac {1}{z^2}-z^2)\geq 0$
and the proof is done

Spoiler

Albert,

You combination of the two given functions into a single objective function is very useful. At this point, you could simply appeal to cyclic symmetry to claim the extrema has to occur for $x=y=z=1$ and alleviate the need for any differentiation.

I realize it is all too easy to come along after the fact and critique the work of others, and I don't mean this as a criticism, but rather as an observation. This observation notwithstanding, your approach is elegant and to be praised. :)