Prove that x^2 +y^2 +z^2-xy-yz-zx is always positive.
Answers
Answered by
93
hey
We have : x2 + y2 + z2 - xy - yz - zy
Now we multiply by in whole equation and get
2 ( x2 + y2 + z2 - xy - yz - zx )
⇒2 x2 + 2y2 + 2z2 - 2xy - 2yz - 2zx
⇒ x2 + x2 + y2 + y2 + z2 + z2 - 2xy - 2yz - 2zx
⇒ x2 + y2 - 2xy + y2 + z2 - 2yz + z2 + x2 - 2zx
⇒ ( x - y )2 + ( y - z )2 + ( z - x )2
From above equation we can see that for distinct value of x , y and z given equation is always positive .
Hope this information will clear your doubts about topic.
We have : x2 + y2 + z2 - xy - yz - zy
Now we multiply by in whole equation and get
2 ( x2 + y2 + z2 - xy - yz - zx )
⇒2 x2 + 2y2 + 2z2 - 2xy - 2yz - 2zx
⇒ x2 + x2 + y2 + y2 + z2 + z2 - 2xy - 2yz - 2zx
⇒ x2 + y2 - 2xy + y2 + z2 - 2yz + z2 + x2 - 2zx
⇒ ( x - y )2 + ( y - z )2 + ( z - x )2
From above equation we can see that for distinct value of x , y and z given equation is always positive .
Hope this information will clear your doubts about topic.
Answered by
81
Answer:
Hello
We have : x2 + y2 + z2 - xy - yz - zy and we have to prove that it is always positive.
Now we multiply the whole equation by 2 and get :
2 ( x² + y² + z² - xy - yz - zx )
⇒2x² + 2y² + 2z² - 2xy - 2yz - 2zx
⇒ x² + x² + y² + y² + z² + z² - 2xy - 2yz - 2zx
⇒ x² + y² - 2xy + y² + z² - 2yz + z² + x² - 2zx
⇒ ( x - y )² + ( y - z )² + ( z - x )²
Step-by-step explanation:
From above equation we can conclude that thegiven equation is always positive because the square of any rational number is always positive
Hope you won't face any more doubt after understanding the above mentioned information.
Thanks....
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