If x^2 + y^2 + z^2 - xy - yz - zx < 0 then which one statement is correct.

M4maths Previous Puzzles

17 Apr 2014 Solve If x^2 + y^2 + z^2 - xy - yz - zx < 0 then which one statement is correct.: OPtion
1) x = y = z
2) x > y > z
3) x < y < z
4) x ≠ y = z
5) x = y ≠ z
6) x ≠ y ≠ z
7) x, y, z have negative values
8) any one of x, y, z is negative
9) any two of x, y, z are negative
10) Not possible; Correct Option: 10 Best Solution (17)

Dr.Dipin Singh 10 years AGO

(x-y)^2 + (y-z)^2 + (z-x)^2 >= 0
2 (x^2 + y^2 + z^2 - xy - yz - zx) >= 0
x^ + y^ + z^2 - xy - yz - zx >= 0

so x^2 + y^2 + z^2 - xy - yz - zx < 0
is not possible.

Like? Yes (4) | No (3)

PAWAN KUMAR 10 years AGO

2*{x^2+y^2+z^2-xy-yz-zx}={(x-y)^2 +(y-z)^2 +(z-x)^2}
As RHS>0, SO, LHS must be >0.

Like? Yes (1) | No

RAKESH 10 years AGO

x^2 + y^2 + z^2 - xy - yz - zx
= 1/2 *[(x-y)^2+(y-z)^2+(z-x)^2] always > 0

so, x^2 + y^2 + z^2 - xy - yz - zx < 0
is not possible.

Like? Yes (5) | No (5)

MP 10 years AGO

(x-y)^2 + (y-z)^2 + (z-x)^2 >= 0
2 (x^2 + y^2 + z^2 - xy - yz - zx) >= 0
x^ + y^ + z^2 - xy - yz - zx >= 0

so x^2 + y^2 + z^2 - xy - yz - zx can never be less than zero.

Like? Yes (3) | No (4)

Ann Theressa 10 years AGO

The equation,

(x - y)^2 + (x - z)^2 + (y - z)^2 >= 0

by virtue of being a sum of squares.
Expanding the equation,

2x^2 + 2y^2 + 2z^2 - 2xy - 2xz - 2yz >= 0.

x^2 + y^2 + z^2 >= xy + xz + yz

From the given equation,
x^2 + y^2 + z^2 - xy - yz - zx < 0
===>
x^2 + y^2 + z^2 < xy + yz + zx
which is not possible

Like? Yes (2) | No (4)

V. Mohan 10 years AGO

not possible

Like? Yes (1) | No (3)

Rashmi bharti 10 years AGO