Prove that x^2+2xy+3y^2-6x-2y >-11
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
We have to prove that x^2 + 2xy + 3y^2 - 6x - 2y > -11
A term in an expression that is a square cannot be negative, therefore:
(x + 2y)^2 + (x - 6)^2 + 2(y - 1)^2 + 38 `>=` 38
An expression greater than 38 is greater than 0
=> (x + 2y)^2 + (x - 6)^2 + 2(y - 1)^2 + 38 `>=` 0
=> x^2 + 4xy + 4y^2 + x^2 - 12x + 36 + 2y^2 - 4y + 2 - 38 `>=` 0
=> 2x^2 + 4xy + 6y^2 - 12x - 4y - 38 `>=` 0
=> x^2 + 2xy + 3y^2 - 6x - 2y - 19 `>=` 0
=> x^2 + 2xy + 3y^2 - 6x - 2y `>=` 19
If an expression is greater than 19 it is greater than -11.
This proves that x^2 + 2xy + 3y^2 - 6x - 2y `>=` -11
knasim | Student | eNotes Newbie
Equivalently we have to prove that 2x2+4xy+6y2-12x-4y+22>=0
From Cauchy Schwarz (CS) Inequality, 4 [(x+2y)2+(y-1)2+(y-1)2+(x-6)2] >= (x+2y+1-y+1-y+6-x)2 = 64
Hence, (x+2y)2+(y-1)2+(y-1)2+(x-6)2 >= 16
or 2x2+4xy+6y2-12x-4y+22>=0
Since the CS equality condition cannot be met here, we have 2x2+4xy+6y2-12x-4y+22>0