MHB Prove x²+y²+z² ≤ 2 For 0 ≤ x,y,z ≤ 1 with xy+yz+zx=1

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The discussion centers on proving the inequality 1 ≤ x² + y² + z² ≤ 2 for variables x, y, and z within the range [0, 1], given the condition xy + yz + zx = 1. Participants share their approaches and solutions, emphasizing the need for a rigorous mathematical proof. The consensus acknowledges the validity of the inequality under the specified conditions. The conversation highlights the collaborative effort in solving the problem, with participants commending each other's contributions. The proof of the inequality is confirmed as a significant mathematical challenge.
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For $x,\,y$ and $z\in [0,\,1]$ such that $xy+yz+zx=1$, prove $1\le x^2+y^2+z^2\le 2$.
 
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anemone said:
For $x,\,y$ and $z\in [0,\,1]$ such that $xy+yz+zx=1$, prove $1\le x^2+y^2+z^2\le 2$.
my solution:
$Using \,\, AP\geq GP$

$left \,\, side\\
x^2+y^2\geq 2xy---(1)\\
y^2+z^2\geq 2yz---(2)\\
z^2+x^2\geq 2zx---(3)\\
(1)+(2)+(3):2(x^2+y^2+z^2)\geq 2(xy+yz+zx)=2\\
\therefore x^2+y^2+z^2\geq 1\\$
$right \,\, side\\
(x+y+z)^2=x^2+y^2+z^2+2\leq x+y+z+2---(4)\\
(for\,\, 0\leq x,y,z\leq 1)\\
let: (x+y+z)=k\\
we \,\, have:k^2-k-2=(k+1)(k-2)\leq 0\rightarrow 0\leq k\leq 2\\
from (4):x^2+y^2+z^2\leq 2$

$the\,\,proof\,\, is \,\, done$
 
Last edited:
Albert said:
my solution:
$Using \,\, AP\geq GP$

$left \,\, side\\
x^2+y^2\geq 2xy---(1)\\
y^2+z^2\geq 2yz---(2)\\
z^2+x^2\geq 2zx---(3)\\
(1)+(2)+(3):2(x^2+y^2+z^2)\geq 2(xy+yz+zx)=2\\
\therefore x^2+y^2+z^2\geq 1\\$
$right \,\, side\\
(x+y+z)^2=x^2+y^2+z^2+2\leq x+y+z+2---(4)\\
(for\,\, 0\leq x,y,z\leq 1)\\
let: (x+y+z)=k\\
we \,\, have:k^2-k-2=(k+1)(k-2)\leq 0\rightarrow 0\leq k\leq 2\\
from (4):x^2+y^2+z^2\leq 2$

$the\,\,proof\,\, is \,\, done$
Good job Albert and thanks for participating!(Cool)
 
anemone said:
For $x,\,y$ and $z\in [0,\,1]$ such that $xy+yz+zx=1$, prove $1\le x^2+y^2+z^2\le 2$.

We have $2x^2+2y^2 + 2z^2 - 2xy - 2yz - 2zx = (x-y)^2 + (y-z)^2 + (z-x)^2$
hence $2x^2+2y^2 + 2z^2 - 2xy - 2yz - 2zx >=0$
or $2x^2+2y^2 + 2z^2 >= 2xy + 2yz + 2zx =2$
or $x^2+y^2+z^2 >= 1$
further
letting $x=\tan\,A$, $y=\tan\, B$, $z\tan\, C$ we have A, B, C, between 0 and $\frac{\pi}{4}$
also $\tan\, A \tan\, B + \tan\, B \tan\, C + \tan\, C \tan\, A = 1$
using $\tan(A+B+C)$ we get
$A+B+C=\frac{\pi}{2}$
so $B+C>=A$ from above and A between 0 and $\frac{\pi}{4}$
so $x <= y+z$
so $x^2 <= xy + zx$
similarly $y^2 <=yz + yx$
and $z^2 <= zx + yz$
adding the 3 we get $x^2+y^2+z^2<=2(xy+yz+zx)$ or 2
 
Last edited:
kaliprasad said:
We have $2x^2+2y^2 + 2z^2 - 2xy - 2yz - 2zx = (x-y)^2 + (y-z)^2 + (z-x)^2$
hence $2x^2+2y^2 + 2z^2 - 2xy - 2yz - 2zx >=0$
or $2x^2+2y^2 + 2z^2 >= 2xy + 2yz + 2zx =2$
or $x^2+y^2+z^2 >= 1$
further
letting $x=\tan\,A$, $y=\tan\, B$, $z\tan\, C$ we have A, B, C, between 0 and $\frac{\pi}{4}$
also $\tan\, A \tan\, B + \tan\, B \tan\, C + \tan\, C \tan\, A = 0$
using $\tan(A+B+C)$ we get
$A+B+C=\frac{\pi}{2}$
so $B+C>=A$ from above and A between 0 and $\frac{\pi}{4}$
so $x <= y+z$
so $x^2 <= xy + zx$
similarly $y^2 <=yz + yx$
and $z^2 <= zx + yz$
adding the 3 we get $x^2+y^2+z^2<=2(xy+yz+zx)$ or 2
Very good job, kaliprasad! Bravo!
 
kaliprasad said:
We have $2x^2+2y^2 + 2z^2 - 2xy - 2yz - 2zx = (x-y)^2 + (y-z)^2 + (z-x)^2$
hence $2x^2+2y^2 + 2z^2 - 2xy - 2yz - 2zx >=0$
or $2x^2+2y^2 + 2z^2 >= 2xy + 2yz + 2zx =2$
or $x^2+y^2+z^2 >= 1$
further
letting $x=\tan\,A$, $y=\tan\, B$, $z\tan\, C$ we have A, B, C, between 0 and $\frac{\pi}{4}$
also $\tan\, A \tan\, B + \tan\, B \tan\, C + \tan\, C \tan\, A = 0$
using $\tan(A+B+C)$ we get
$A+B+C=\frac{\pi}{2}$
so $B+C>=A$ from above and A between 0 and $\frac{\pi}{4}$
so $x <= y+z$
so $x^2 <= xy + zx$
similarly $y^2 <=yz + yx$
and $z^2 <= zx + yz$
adding the 3 we get $x^2+y^2+z^2<=2(xy+yz+zx)$ or 2
$\tan\, A \tan\, B + \tan\, B \tan\, C + \tan\, C \tan\, A =0$ ,$why ?$
 
Albert said:
$\tan\, A \tan\, B + \tan\, B \tan\, C + \tan\, C \tan\, A =0$ ,$why ?$

it was a typo I have done corrections should be 1
 
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