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

[anemone]

For $x,\,y$ and $z\in [0,\,1]$ such that $xy+yz+zx=1$, prove $1\le x^2+y^2+z^2\le 2$.

 

[Albert1]

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:

Spoiler

$Using \,\, AP\geq GP$

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

$the\,\,proof\,\, is \,\, done$

 

[anemone]

my solution:

Spoiler

$Using \,\, AP\geq GP$

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

$the\,\,proof\,\, is \,\, done$

Good job Albert and thanks for participating!(Cool)

 

[kaliprasad]

For $x,\,y$ and $z\in [0,\,1]$ such that $xy+yz+zx=1$, prove $1\le x^2+y^2+z^2\le 2$.

Spoiler

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

 

[anemone]

Spoiler

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!

 

[Albert1]

Spoiler

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

Spoiler

$\tan\, A \tan\, B + \tan\, B \tan\, C + \tan\, C \tan\, A =0$ ,$why ?$

 

[kaliprasad]

Spoiler

$\tan\, A \tan\, B + \tan\, B \tan\, C + \tan\, C \tan\, A =0$ ,$why ?$

Spoiler

it was a typo I have done corrections should be 1