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

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For positive real numbers x, y, and z satisfying x+y+z=3, it is proven that 1/x² + 1/y² + 1/z² is greater than or equal to x² + y² + z². The discussion focuses on establishing this inequality through various mathematical approaches. Participants explore potential proofs and share insights on the properties of these expressions. The conversation emphasizes the importance of conditions on x, y, and z in deriving the inequality. Ultimately, the proof highlights a significant relationship between the sums of squares and their reciprocals.
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Let $x,\,y,\,z$ be positive real numbers such that $x+y+z=3$.

Prove that $\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\ge x^2+y^2+z^2$.
 
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anemone said:
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:
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
 
Albert said:
my solution:
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

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. :)
 
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