prove that for all real x,y the value of x2 + 2xy + 3y2 - 6x - 2y cannot be less than -11.
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2 Answers
491/2(2x2+4xy+6y2-12x-4y)
=1/2(x2+2.x.2y+4y2 +2y2-2.√2y.√2+2+x2-2.x.6+36-38)
=1/2(x+2y)2+1/2(√2y-√2)2+1/2(x-6)2-19
i am getting least value as -19
don't know if i am wrong...i mean in some calculations...
1take this equation as
x^2 + 2xy + 3y^2 - 6x - 2y + 11 \geq 0
\Rightarrow x^2 + 2x(y - 3) + 3y^2 - 2y + 11 \geq 0
here D \geq 0
after solving this
we will get
\Rightarrow (y-1)^2 \geq 0
which is true for every value of y
so the given condition is satisfied
