my solution:anemone said:For $x,\,y$ and $z\in [0,\,1]$ such that $xy+yz+zx=1$, prove $1\le x^2+y^2+z^2\le 2$.
Good job Albert and thanks for participating!(Cool)Albert said:my solution:
$Using \,\, AP\geq GP$
$left \,\, side\\
x^2+y^2\geq 2xy---(1)\\
y^2+z^2\geq 2yz---(2)\\
z^2+x^2\geq 2zx---(3)\\
(1)+(2)+(3):2(x^2+y^2+z^2)\geq 2(xy+yz+zx)=2\\
\therefore x^2+y^2+z^2\geq 1\\$
$right \,\, side\\
(x+y+z)^2=x^2+y^2+z^2+2\leq x+y+z+2---(4)\\
(for\,\, 0\leq x,y,z\leq 1)\\
let: (x+y+z)=k\\
we \,\, have:k^2-k-2=(k+1)(k-2)\leq 0\rightarrow 0\leq k\leq 2\\
from (4):x^2+y^2+z^2\leq 2$
$the\,\,proof\,\, is \,\, done$
anemone said:For $x,\,y$ and $z\in [0,\,1]$ such that $xy+yz+zx=1$, prove $1\le x^2+y^2+z^2\le 2$.
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
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
Albert said:$\tan\, A \tan\, B + \tan\, B \tan\, C + \tan\, C \tan\, A =0$ ,$why ?$
The inequality x²+y²+z² ≤ 2 for 0 ≤ x,y,z ≤ 1 is a constraint that limits the values of x, y, and z within a defined range. This inequality is important in many mathematical and scientific applications, as it helps to define a restricted region in three-dimensional space.
The given condition of xy+yz+zx=1 can be rewritten as (x+y+z)²-2xyz=1. This, combined with the fact that 0 ≤ x,y,z ≤ 1, allows us to make the substitution x²+y²+z²=2-2xyz. By substituting this into the original inequality, we get 2-2xyz ≤ 2, which simplifies to xyz ≥ 0. This means that the values of x, y, and z must all be positive or negative, which is in line with the given conditions of 0 ≤ x,y,z ≤ 1.
To prove the inequality x²+y²+z² ≤ 2 for 0 ≤ x,y,z ≤ 1 with xy+yz+zx=1, we can use the method of Lagrange multipliers. By setting up the Lagrangian function L(x,y,z,λ) = x²+y²+z²-λ(xy+yz+zx-1) and solving for the critical points, we can show that the maximum value of x²+y²+z² occurs when x=y=z=1, which satisfies the given conditions and results in a value of 2 for x²+y²+z². Therefore, the inequality is proven.
Yes, the inequality x²+y²+z² ≤ 2 can be generalized to higher dimensions. In fact, it is a special case of the more general inequality x₁²+x₂²+...+xₙ² ≤ n, where x₁,x₂,...,xₙ are variables and n is a positive real number. This inequality defines a hypersphere in n-dimensional space and is useful in many mathematical and physical applications.
The inequality x²+y²+z² ≤ 2 is closely related to the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. This can be seen by setting x=a/c, y=b/c, z=0, where a, b, and c are the side lengths of a right triangle. This relationship between the inequality and the Pythagorean theorem can be extended to higher dimensions as well.