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

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24 May 2016 Correct statement

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

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

Solution

x^2 + y^2 + z^2 – xy - yz – zx < 0
=> 2x^2 + 2y^2 + 2z^2 – 2xy - 2yz – 2zx < 0
=> x^2 + y^2 - 2xy + y^2 + z^2 - 2yz + z^2 + x^2 - 2zx < 0
=> (x – y)^2 + (y – z)^2 + (z – x)^2 < 0
(x – y)^2 + (y – z)^2 + (z – x)^2 ,
This expression is the sum of three perfect square numbers and it is not possible to make this expression less than 0.

Option 10)

Correct Option: 10 Best Solution (15)

Aakash Beniwal   8 years AGO

if x=y=z then solution = 0, in all other case solution is greater than 0.
Hence solution is : Not possible

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ABHINAV PUSHPAM   8 years AGO

if x=y=z, then the equation becomes 0, in all other cases it will be more than 0. So it is not possible..

Option 10)

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RAKESH   8 years AGO

given exp is eq to = 1/2 *[(x-y)^2+(y-z)^2+(z-x)^2]
alwz +ve for any x,y,z

Like? Yes (1) | No   

RAHUL KUMAR   8 years AGO

Put some arbitrary values for x, y, z and evaluate the expression. None of them seems to yield a negative answer.

For example,
1. Put x=y=z=1
2. Put x=3, y=2, z=1
3. Put x=1, y=2, z=3
4. Put x=1, y=z=2
5. Put x=y=1, z=2
6. Put x=1, y=2, z=3
7. Put x=-1, y=-2, z=-3
8. Put x=-1, y=2, z=3
9. Put x=-1, y=-2, z=3
Every-time, x2 + y2 + z2 - xy -yz - xz yields a non-negative number.
Hence, the result not follows.

option (10)

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Akshat Sood   8 years AGO

It is not possible as the term contains square terms
if we take negative terms though we will get +ve answer.

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viswambher   8 years AGO

Use AM-GM inequality.

x^2 + y^2 >(or equal) 2xy
y^2 + z^2 > 2yz
z^2 + x^2 > 2xz

Adding all the equations.

x^2+y^2+z^2-xy-yz-xz >(or equal) 0

So the given equation is always greater than or equal to zero but never less than zero

Alternately, we can write

x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx)

from this , the given equation must be greater than or equal to zero

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Mukund Manohar   8 years AGO

x^2+y^2+z^2-xy-yz-zx < 0
=> 2x^2+2y^2+2z^2-2xy-2yz-2zx < 0
=> (x^2+y^2-2xy)+(y^2+z^2-2yz)+(x^2+z^2-2zx) < 0
=> (x-y)^2 + (y-z)^2 + (x-z)^2 < 0
It is not possible that the sum of 3 perfect square is less than 0

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Sivasis dash   8 years AGO

x^2+y^2+z^2-xy-yz-zx1/2[2x^2+2y^2+2z^2-2xy-2yz-2zx][(x-y)^2+(y-z)^2+(z-x)^2]

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shivani   8 years AGO

bcz this expression can be reduced to the sum of perfect square so it can not be less than zero

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Siva Seshadri C   8 years AGO

x^2+y^2+z^2< xy+yz+zx
for all the options inequqlity nt satisfied

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Rohit Kumar   8 years AGO

Put some arbitrary values for x,
y, z and evaluate the
expression. None of them seems
to yield a negative answer.
For example,
1. Put x=y=z=1
2. Put x=3, y=2, z=1
3. Put x=1, y=2, z=3
4. Put x=1, y=z=2
5. Put x=y=1, z=2
6. Put x=1, y=2, z=3
7. Put x=-1, y=-2, z=-3
8. Put x=-1, y=2, z=3
9. Put x=-1, y=-2, z=3
Every-time, x2 + y2 + z2 - xy -yz
- xz yields a non-negative
number.
Hence, the result follows.

Like? Yes | No   

Kunal nemu   8 years AGO

if x=y=z then the equation is become 0 ,in all other cases it become more than 0, so it is not possible
so answer is option(10)

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Rakhesh Gurle   8 years AGO

try with x,y,z equals to 1,2,3 and then verify all the options with different combinations of x,y,z then you will find option 10 as answer

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ankusha   8 years AGO

x^2+y^2+z^2-xy-yz-zx=((x-y)^2 + (y-z)^2 + (z-y)^2) /2;
which is sum of squares......
cannot be < 0.

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shanmuksrinivas   8 years AGO

because at all conditions it give positive integer ,so i cant happen

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